Statistical Inference

Problem 1101

Public health officials believe that 90.6% of children have been vaccinated against measles. A random survey of medical records at many schools across the country found that, among more than 13,000 children, only 89.8% had been vaccinated. A statistician would reject the 90% hypothesis with a P-value of P=0.022P = 0.022.
a) Explain what the P-value means in this context. b) The result is statistically significant, but is it important? Comment.
A. We concluded that the actual percentage of vaccinated children is below 90.6%. A 0.8% drop would probably not be considered noteworthy but in context, if 1,000,000 children are vaccinated each year a 0.8% difference accounts for 8000 more children not being vaccinated, which is important.
B. A 0.8% difference in child vaccinations in not important.
C. We conclude that the actual percentage of vaccinated children is below 90.6% and is about 89.8%. This drop is not important because only a 5% change or more can be considered important.

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Problem 1102

Back in 1993, a study found 52 in a sample of 503 European women were mathematicians. A similar study, conducted in 2005, found 88 in a sample of 495 European women were mathematicians. Let p1p_1, p2p_2 represent the proportion of European woman mathematicians in 1993 and in 2005 respectively. At significance level 0.002, is this good evidence showing that the proportion of women mathematicians in Europe has changed in 12 years since 1993?
A) State the Null and the Alternate Hypotheses: H0:p1=p2H_0: p_1 = p_2 (dp=0dp = 0) H1:p1p2H_1: p_1 \ne p_2 (dp0dp \ne 0)
B) Is this a (left-tail, right-tail, one-tail, two-tail) test? two-tail
C) Using 4 decimal places for p1p_1 and p2p_2, find the value of dp^d\hat{p}. dp^=d\hat{p} = ____ Note: dp^d\hat{p} must be in 4 decimal places.
D) Find the value of p^c\hat{p}_c. p^c=1.1403\hat{p}_c = 1.1403 Round your answer to 4 decimal places.
E) Find σdp^\sigma_{d\hat{p}}, the standard error of dp^d\hat{p}'s: σdp^=\sigma_{d\hat{p}} = ____ Round your answer to 4 decimal places.
F) Calculate the z-score of your sampling difference in proportion. zdp^=z_{d\hat{p}} = ____ Enter your answer in 2 decimal places.
G) Using z-score from part (F), what is the p-value of this test? p-value == ____

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Problem 1103

etermine whether the study depicts an observational study or an experiment. poll is conducted by a school's English department in which seventh-grade students are asked f they prefer to be in their English class or their math class.
Choose the correct description of the study. A. The study is an observational study because the researchers control one variable to determine the effect on the response variable. B. The study is an experiment because the researchers control one variable to determine the effect on the response variable. C. The study is an observational study because the study examines individuals in a sample, but does not try to influence the response variable. D. The study is an experiment because the study examines individuals in a sample, but does not try to influence the variable of interest.

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Problem 1104

38 married couples without children are asked to report the number of times per year they initiate a date night. The men report initiating an average of 8.3 date nights with a standard deviation of 4.1, though it's possible they're overstating to make themselves look good. Is there significant evidence to conclude that married men without children initiate date night 7 times per year at the 0.05 significance level? Note that there's evidence that this distribution is skewed.
What are the hypotheses? H0:μ7H_0: \mu \le 7 vs H1:μ>7H_1: \mu > 7 H0:μ=7H_0: \mu = 7 vs H1:μ7H_1: \mu \ne 7 H0:μ=8.3H_0: \mu = 8.3 vs H1:μ8.3H_1: \mu \ne 8.3 H0:μ8.3H_0: \mu \le 8.3 vs H1:μ>8.3H_1: \mu > 8.3
What distribution does the test statistic follow? t with 39 degrees of freedom t with 37 degrees of freedom t with 38 degrees of freedom z
What is the value of the test statistic? Round to two decimal places. 1.96
What is probability statement for the p-value? 2P(Ttest statistic)2P(T \ne \text{test statistic}) P(T>test statistic)P(T > \text{test statistic}) 2P(Ztest statistic)2P(Z \ne \text{test statistic}) P(Ztest statistic)P(Z \ge \text{test statistic}) None of the above

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Problem 1105

QUESTION 12 Provide an appropriate response. Find the value of EE, the margin of error, for c=0.90,n=10c=0.90, n=10 and s=3.7s=3.7. 0.68 2.14 1.62 2.12

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Problem 1106

Determine whether the given value is a statistic or a parameter. A sample of students is selected and it is found that 55% own a television.
Choose the correct statement below. Parameter Statistic

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Problem 1107

A null and alternative hypotheses are given. Determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed. H0:σ=9H_0: \sigma = 9 H1:σ<9H_1: \sigma < 9
What type of test is being conducted in this problem? Left-tailed test Right-tailed test Two-tailed test

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Problem 1108

For the following data set, compute the correlation coefficient. \begin{tabular}{c|ccccc} x & 2 & 4 & 8 & 8 & 9 \\ \hline y & 1.4 & 1.5 & 2.1 & 1.9 & 2.5 \end{tabular}
Formula sheet rr \approx \square (Round to four decimal places as needed.)

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Problem 1109

Several years ago, the mean height of women 20 years of age or older was 63.7 inches. Suppose that a random sample of 45 women who are 20 years of age or older today results in a mean height of 64.3 inches. (a) State the appropriate null and alternative hypotheses to assess whether women are taller today. (b) Suppose the P -value for this test is 0.13 . Explain what this value represents. (c) Write a conclusion for this hypothesis test assuming an α=0.10\alpha=0.10 level of significance. (a) State the appropriate null and alternative hypotheses to assess whether women are taller today. A. H0:μ=64.3H_{0}: \mu=64.3 in. versus H1:μ>64.3H_{1}: \mu>64.3 in. B. H0:μ=63.7H_{0}: \mu=63.7 in. versus H1:μ63.7inH_{1}: \mu \neq 63.7 \mathrm{in}. C. H0:μ=63.7H_{0}: \mu=63.7 in. versus H1:μ>63.7H_{1}: \mu>63.7 in. D. H0:μ=64.3H_{0}: \mu=64.3 in. versus H1:μ64.3inH_{1}: \mu \neq 64.3 \mathrm{in}. E. H0:μ=63.7H_{0}: \mu=63.7 in. versus H1:μ<63.7inH_{1}: \mu<63.7 \mathrm{in}. F. H0:μ=64.3H_{0}: \mu=64.3 in. versus H1:μ<64.3H_{1}: \mu<64.3 in.

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Problem 1110

If consumers spend 80 cents out of every extra dollar received, the
Multiple Choice
multiplier is 5.
marginal propensity to save is 0.80.
marginal propensity to consume is 0.20.
multiplier is 20.

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Problem 1111

Use the given data to complete parts (a) and (b) below.
x | y ---|--- 2.3 | 4 3.9 | 1.5 2.8 | 3.6 4.7 | 4.9
Compute the linear correlation coefficient with the additional data point. The linear correlation coefficient for the five pieces of data is \boxed{}. (Round to three decimal places as needed.)
Comment on the effect the additional data point has on the linear correlation coefficient.
A. The additional data point does not affect the linear correlation coefficient. B. The additional data point strengthens the appearance of a linear association between the data points. C. The additional data point weakens the appearance of a linear association between the data points.
Explain why correlations should always be reported with scatter diagrams.
A. The scatter diagram can be used to distinguish between association and causation. B. The scatter diagram is needed to see if the correlation coefficient is being affected by the presence of outliers.

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Problem 1112

n=2317n = 2317 x=411x = 411 A survey of 2317 adults in a certain large country aged 18 and older conducted by a reputable polling organization found that 411 have donated blood in the past two years. Complete parts (a) through (c) below.
(a) Obtain a point estimate for the population proportion of adults in the country aged 18 and older who have donated blood in the past two years. p^=\hat{p} = (Round to three decimal places as needed.)
(b) Verify that the requirements for constructing a confidence interval about pp are satisfied. The sample a simple random sample, the value of is , which is 10, and the less than or equal to 5% of the . (Round to three decimal places as needed.)
(c) Construct and interpret a 90% confidence interval for the population proportion of adults in the country who have donated blood in the past two years. Select the correct choice below and fill in any answer boxes within your choice. (Type integers or decimals rounded to three decimal places as needed. Use ascending order.) A. We are % confident the proportion of adults in the country aged 18 and older who have donated blood in the past two years is between and . B. There is a % chance the proportion of adults in the country aged 18 and older who have donated blood in the past two years is between and .

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Problem 1113

A sample of size n=66n = 66 is drawn from a normal population whose standard deviation is σ=6.2\sigma = 6.2. The sample mean is x=37.77\overline{x} = 37.77.
Part: 0 / 2
Part 1 of 2
(a) Construct a 99% confidence interval for μ\mu. Round the answer to at least two decimal places.
A 99% confidence interval for the mean is 00<μ<00\boxed{\phantom{00}} < \mu < \boxed{\phantom{00}}.

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Problem 1114

Sleeping outlier: A simple random sample of eight college freshmen were asked how many hours of sleep they typically got per night. The results were 8.5 24 8.5 8 7.5 9 6.5 6 Send data to Excel Notice that one joker said that he sleeps 24 a day. Part: 0 / 3 Part 1 of 3 (a) The data contain an outlier that is clearly a mistake. Eliminate the outlier, then construct a 99%99\% confidence interval for the mean amount of sleep from the remaining values. Round the answers to at least two decimal places. A 99%99\% confidence interval for the mean amount of sleep from the remaining values is \boxed{} < μ\mu < \boxed{}.

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Problem 1115

En el muestreo aleatorio simple la muestra se selecciona a través de un sistema que se aplica después de la selección de un punto aleatorio d Falso Verdadero

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Problem 1116

Español ek and Stone Mountain are schools in different states. The heights of students at Castle Creek have a population mean of 59.7 inches and a standard of 1.4 inches. The heights of students at Stone Mountain have a population mean of 56.7 inches with a standard deviation of 1.5 inches. For each distribution of the heights of students is clearly bell-shaped. student at Castle Creek and is 57 inches tall. Melissa is a student at Stone Mountain and is 53 inches tall. d the zz-scores of Maya's height as a student at Castle Creek and Melissa's height as a student at Stone Mountain. und your answers to two decimal places. -score of Maya's height: \square -score of Melissa's height: \square ative to her population, which student is taller? ose the best answer based on the zz-scores of the two heights.
Maya Melissa It is unclear which student is taller relative to her population

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Problem 1117

Researchers wanting to assess level of heel pain associated with a new footwear developed for those with plantar fasciitis. They randomly sample 120 people with a history of plantar fasciitis and 120 people without a history of plantar fasciitis and ask them to wear the shoes and report the level of heel pain (1-10) they experience after walking in the shoes for 2 hours. The average heel pain score for the group with plantar fasciitis was 1.4 and the average heel pain score for the group without plantar fasciitis was 1.2. The Levene's test for equality of variances had a pp value of 0.11. You know this means: The researchers should report the tt-test results for assuming equal variances The researchers should report the t-test results NOT assuming equal variances The researchers should reject the null hypothesis and report there is a difference in the average heel pain score The researchers should fail to reject the null hypothesis and conclude there is NOT a difference in the average heel pain score

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Problem 1118

Ihis question: 1 point(S) possible A data set includes 109 body temperatures of healthy adult humans having a mean of 98.2F98.2^{\circ} \mathrm{F} and a standard deviation of 0.62F0.62^{\circ} \mathrm{F}. Construct a 99%99 \% confidence interval estimate of the mean body temperatu humans. What does the sample suggest about the use of 986F986^{\circ} \mathrm{F} as the mean body temperature? Click here to view a t distribution table. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table.
What is the confidence interval estimate of the population mean μ\mu ? \square F<μ<{ }^{\circ} \mathrm{F}<\mu< \square F{ }^{\circ} \mathrm{F} (Round to three decimal places as needed.) What does this suggest about the use of 986F986^{\circ} \mathrm{F} as the mean body temperature? A. This suggests that the mean body temperature is higher than 98.6F98.6^{\circ} \mathrm{F} B. This suggests that the mean body temperature is lower than 98.6F98.6^{\circ} \mathrm{F} C. This suggests that the mean body temperature could very possibly be 98.6F98.6^{\circ} \mathrm{F}

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Problem 1119

Find the standardized test statistic t for a sample with n=10n = 10, xˉ=8.8\bar{x} = 8.8, s=1.3s = 1.3, and α=0.05\alpha = 0.05 if H0:μ9.7H_0: \mu \le 9.7. Round your answer to three decimal places.
A. 2.189 -2.189 B. 2.617 -2.617 C. 3.010 -3.010 D. 3.186 -3.186

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Problem 1120

Which statement best reflects the relationship between correlation and causation? A, B, C, or D?

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Problem 1121

Is the cost of fabric types proportional? Fabric A: \$31.25 for 5 sq. yd, Fabric B: \$71.50 for 11 sq. yd. Explain.

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Problem 1122

Previous Problem Problem List Next Problem (1 point) (Exam FM Sample Problem 199) Miaoqi and Nui entered into a six year interest rate swap on May 5, 2015. The notional amount of the swap was a level 30000 for all six years. The swap has annual settlement periods with the first period starting on May 5, 2015. Under the swap, Miaoqi agreed to pay a variable rate based on the one year spot rate at the beginning of each settlement period. Nui will pay Miaoqi the fixed rate of 7.6%7.6 \% on each settlement date. On May 5, 2017, the spot interest rate curve was as follows: \begin{tabular}{|c|c|c|c|c|c|c|} \hline Time & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline Spot Rate 6.5%6.5 \% & 7%7 \% & 7.5%7.5 \% & 7.75%7.75 \% & 8.25%8.25 \% & 8.5%8.5 \% \\ \hline \end{tabular}
Miaoqi decides that she wants to sell the swap on May 5, 2017. Calculate the market value of the swap on May 5, 2017, from Miaoqids position in the swap.

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Problem 1123

Previous Problem Problem List Next Problem (1 point) (Exam FM Sample Problem 199) Miaoqi and Nui entered into a six year interest rate swap on May 5, 2015. The notional amount of the swap was a level 30000 for all six years. The swap has annual settlement periods with the first period starting on May 5, 2015. Under the swap, Miaoqi agreed to pay a variable rate based on the one year spot rate at the beginning of each settlement period. Nui will pay Miaoqi the fixed rate of 7.6%7.6 \% on each settlement date. On May 5, 2017, the spot interest rate curve was as follows: \begin{tabular}{|c|c|c|c|c|c|c|} \hline Time & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline Spot Rate 6.5%6.5 \% & 7%7 \% & 7.5%7.5 \% & 7.75%7.75 \% & 8.25%8.25 \% & 8.5%8.5 \% \\ \hline \end{tabular}
Miaoqi decides that she wants to sell the swap on May 5, 2017. Calculate the market value of the swap on May 5, 2017, from Miaoqids position in the swap.

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Problem 1124

A bolt manufacturer is very concerned about the consistency with which his machines produce bolts. The bolts should be 0.23 centimeters in diameter. The variance of the bolts should be 0.015 . A random sample of 29 bolts has an average diameter of 0.22 cm with a standard deviation of 0.0742 . Can the manufacturer conclude that the bolts vary from the required variance at α=0.05\alpha=0.05 level?
Step 1 of 5: State the hypotheses in terms of the standard deviation. Round the standard deviation to four decimal places when necessary.
Answer 2 Points Tables Keypad Keyboard Shortcuts H0:Ha:\begin{array}{l} H_{0}: \square \\ H_{a}: \square \end{array}

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Problem 1125

xx | 1 | 2 | 3 | 4 | 5 | 6 ---|---|---|---|---|---|---| yy | 849 | 1351 | 2108 | 3224 | 4731 | 8157
Use exponential regression to find an exponential function that best fits this data.
f(x)=f(x) =
Use linear regression to find a linear function that best fits this data.
g(x)=g(x) =
Of these two, which equation best fits the data?
Linear
Exponential

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Problem 1126

Refer to the accompanying data display that results from a sample of airport data TInterval speeds in Mbps. Complete parts (a) through (c) below.
Click the icon to view a t distribution table. (13.046,22.15)x=17.598Sx=16.01712719n=50\begin{array}{l} (13.046,22.15) \\ x=17.598 \\ S x=16.01712719 \\ n=50 \end{array} a. What is the number of degrees of freedom that should be used for finding the critical value tα/2t_{\alpha / 2} ? df=\mathrm{df}= (Type a whole number.)

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Problem 1127

Find the P-value for the indicated hypothesis test with the given standardized test statistic, zz. Decide whether to reject H0H_0 for the given level of significance α\alpha. Right-tailed test with test statistic z=1.39z = 1.39 and α=0.04\alpha = 0.04. P-value = (Round to four decimal places as needed.)

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Problem 1128

1. Consider a normal population distribution with the value of σ\sigma known. a. What is the confidence level for the interval xˉ±2.81σ/n\bar{x} \pm 2.81 \sigma / \sqrt{n}? b. What is the confidence level for the interval xˉ±1.44σ/n\bar{x} \pm 1.44 \sigma / \sqrt{n}? c. What value of zα/2z_{\alpha/2} in the CI formula (7.5) results in a confidence level of 99.7%? d. Answer the question posed in part (c) for a confidence level of 75%.

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Problem 1129

4. A CI is desired for the true average stray-load loss μ\mu (watts) for a certain type of induction motor when the line current is held at 10 amps for a speed of 1500 rpm. Assume that stray-load loss is normally distributed with σ=3.0\sigma = 3.0.
a. Compute a 95% CI for μ\mu when n=25n = 25 and xˉ=58.3\bar{x} = 58.3.
b. Compute a 95% CI for μ\mu when n=100n = 100 and xˉ=58.3\bar{x} = 58.3.
c. Compute a 99% CI for μ\mu when n=100n = 100 and xˉ=58.3\bar{x} = 58.3.
d. Compute an 82% CI for μ\mu when n=100n = 100 and xˉ=58.3\bar{x} = 58.3.
e. How large must nn be if the width of the 99% interval for μ\mu is to be 1.0?

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Problem 1130

12. A random sample of 110 lightning flashes in a certain region resulted in a sample average radar echo duration of 0.810.81 sec and a sample standard deviation of 0.340.34 sec ("Lightning Strikes to an Airplane in a Thunderstorm," J. of Aircraft, 1984: 607-611). Calculate a 99% (two-sided) confidence interval for the true average echo duration μ\mu, and interpret the resulting interval.

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Problem 1131

13. The article "Extravisual Damage Detection: Defining the Standard Normal Tree" (Photogrammetric Engr. and Remote Sensing, 1981: 515-522) discusses there use of color infrared photography in identification of normal trees in Douglas fir stands. Among data reported were summary statistics for green-filter analytic optical densitometric measurements on samples of both healthy and diseased trees. For a sample of 69 healthy trees, the sample mean dye-layer density was 1.028, and the sample standard deviation was 0.163. a. Calculate a 95%95\% (two-sided) CI for the true average dye-layer density for all such trees. b. Suppose the investigators had made a rough guess of 0.16 for the value of ss before collecting late a 95%95\% (two-sided) confidence interval for the proportion of all dies that pass the probe.

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Problem 1132

Concentration of CO2CO_2 in the Atmosphere
Levels of carbon dioxide (CO2CO_2) in the atmosphere are rising rapidly, far above any levels ever before recorded. Levels were around 280 parts per million in 1800, before the Industrial Age, and had never, in the hundreds of thousands of years before that, gone above 300 ppm. Levels are now over 400 ppm. The table below shows the rapid rise of CO2CO_2 concentrations over the 55 years from 1960-2015 (data also available in Carbon Dioxide). We can use this information to predict CO2CO_2 levels in different years.
| Year | CO2CO_2 | |---|---| | 1960 | 316.91 | | 1965 | 320.04 | | 1970 | 325.48 | | 1975 | 331.11 | | 1980 | 338.75 | | 1985 | 346.12 | | 1990 | 354.39 | | 1995 | 360.82 | | 2000 | 369.55 | | 2005 | 379.80 | | 2010 | 389.90 | | 2015 | 400.81 |
Concentration of carbon dioxide in the atmosphere
Click here to the dataset associated with this question. Use the 3 e version of the dataset.
If using StatKey, the data needed is preloaded in the drop-down menu in the upper left corner. Click here to access StatKey.
Dr. Pieter Tans, NOAA/ESRL, http://www.esrl.noaa.gov/gmd/ccgg/trends/. Values recorded at the Mauna Loa Observatory in Hawaii.
(a) What is the explanatory variable? What is the response variable? * Year is the explanatory variable and CO2CO_2 concentration is the response variable. CO2CO_2 concentration is the explanatory variable and Year is the response variable.
(b) Use technology to find the correlation between year and CO2CO_2 levels. Round your answer to three decimal places.
(c) Use technology to calculate the regression line to predict CO2CO_2 from year. Round your answer for the intercept to one decimal place and your answer for the slope to three decimal places. CO2=CO_2 = ____ + ____ (Year)
(d) Interpret the slope of the regression line, in terms of carbon dioxide concentrations. The slope tells the predicted number of years for the CO2CO_2 level to go up by that amount. The slope tells the predicted number of years for the CO2CO_2 level to go up by one. The slope tells the predicted CO2CO_2 level one year later. The slope tells the predicted change in CO2CO_2 level one year later.
(e) What is the intercept of the line? Round your answer to one decimal place. The intercept is ____ (Does it make sense in context?)
(f) Use the regression line to predict the CO2CO_2 level in 2003. Use rounded slope and the intercept from part (c). Then round your answer to one decimal place. CO2CO_2 level in 2003: ____
Use the regression line to predict the CO2CO_2 level in 2005. Use rounded slope and the intercept from part (c). Then round your answer to one decimal place. CO2CO_2 level in 2005: ____
(g) Find the residual for 2010. Use rounded slope and the intercept from part (c). Then round your answer to two decimal places. Residual for 2010: ____

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Problem 1133

Computer output for fitting a simple linear model is given below. State the value of the sample slope for this model and give the null and alternative hypotheses for testing if the slope in the population is different from zero. Identify the pp-value and use it (and a 5\% significance level) to make a clear conclusion about the effectiveness of the model.
The regression equation is Y=84.80.0127XY = 84.8 - 0.0127X.
Predictor | Coef | SE Coef | T | P ---|---|---|---|---| Constant | 84.81 | 12.17 | 6.97 | 0.000 X | -0.01265 | 0.01054 | -1.20 | 0.245
Sample slope: pp-value:
Does X appear to be an effective predictor of the response variable Y? eTextbook and Media Save for Later Attempts: 0 of 3 used

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Problem 1134

A potato chip manufacturer produces bags of potato chips that are supposed to have a net weight of 326 grams. Because the chips vary in size, it is difficult to fill the bags to the exact weight desired. However, the bags pass inspection so long as the standard deviation of their weights is no more than 4 grams. A quality control inspector wished to test the claim that one batch of bags has a standard deviation of more than 4 grams, and thus does not pass inspection. If a sample of 28 bags of potato chips is taken and the standard deviation is found to be 4.6 grams, does this evidence, at the 0.025 level of significance, support the claim that the bags should fail inspection? Assume that the weights of the bags of potato chips are normally distributed.
Step 3 of 3: Draw a conclusion and interpret the decision.
Answer 2 Points Tables Keypad Keyboard Shortcuts
We reject the null hypothesis and conclude that there is sufficient evidence at a 0.025 level of significance that the bags should fail inspection.
We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.025 level of significance that the bags should fail inspection.
We fail to reject the null hypothesis and conclude that there is sufficient evidence at a 0.025 level of significance that the bags should fail inspection.
We reject the null hypothesis and conclude that there is insufficient evidence at a 0.025 level of significance that the bags should fail inspection.

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Problem 1135

Pollution and altitude: In a random sample of 331 cars driven at low altitudes, 52 of them exceeded a standard of 11 grams of particulate pollution per gallon of fuel consumed. In an independent random sample of 85 cars driven at high altitudes, 23 of them exceeded the standard. Can you conclude that the proportion of highaltitude vehicles exceeding the standard differs from the proportion of lowaltitude vehicles exceeding the standard? Let p1 denote the proportion of lowaltitude vehicles exceeding the standard and p2 denote the proportion of highaltitude vehicles exceeding the standard. Use the α=0.01 level of significance and the Pvalue method with the TI84 Plus calculator.Pollution\ and\ altitude:\ In\ a\ random\ sample\ of\ 331\ cars\ driven\ at\ low\ altitudes,\ 52\ of\ them\ exceeded\ a\ standard\ of\ 11\ grams\ of\ particulate\ pollution\ per\ gallon\ of\ fuel\ consumed.\ In\ an\ independent\ random\ sample\ of\ 85\ cars\ driven\ at\ high\ altitudes,\ 23\ of\ them\ exceeded\ the\ standard.\ Can\ you\ conclude\ that\ the\ proportion\ of\ high-altitude\ vehicles\ exceeding\ the\ standard\ differs\ from\ the\ proportion\ of\ low-altitude\ vehicles\ exceeding\ the\ standard?\ Let\ p_1\ denote\ the\ proportion\ of\ low-altitude\ vehicles\ exceeding\ the\ standard\ and\ p_2\ denote\ the\ proportion\ of\ high-altitude\ vehicles\ exceeding\ the\ standard.\ Use\ the\ \alpha = 0.01\ level\ of\ significance\ and\ the\ P-value\ method\ with\ the\ TI-84\ Plus\ calculator.
Part: 0/4Part:\ 0/4
Part 1 of 4Part\ 1\ of\ 4
State the appropriate null and alternate hypotheses.State\ the\ appropriate\ null\ and\ alternate\ hypotheses.
H0:H_0:
H1:H_1:
This hypothesis test is a (Choose one) test.This\ hypothesis\ test\ is\ a\ (Choose\ one)\ test.

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Problem 1136

3. Tricia surveys students in her computer class about time spent on computers by students in her school. Will the survey results from this sample support a valid inference? Explain.

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Problem 1137

A variable x\boldsymbol{x} is normally distributed with mean 25 and standard deviation 3. Round your answers to the nearest hundredth as needed. a) Determine the zz-score for x=28x=28. z=z= \square b) Determine the zz-score for x=20x=20. z=z= \square c) What value of xx has a zz-score of 1.33 ? x=x= \square d) What value of xx has a zz-score of 0.3 ? x=x= \square e) What value of xx has a zz-score of 0 ? x=x= \square

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Problem 1138

The lengths of mature trout in a local lake are approximately normally distributed with a mean of μ=13.7\mu=13.7 inches, and a standard deviation of σ=1.6\sigma=1.6 inches.
Fill in the indicated boxes.
Find the z -score corresponding to a fish that is 13.3\mathbf{1 3 . 3} inches long. Round your answer to the nearest hundredth as needed. z=z= \square How long is a fish that has a z -score of 0.2 ? Round your answer to the nearest tenth as needed. \square inches

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Problem 1139

Apply Jefferson's method to the following sets of states with the given populations. Assume that 100 delegates are to be apportioned. State whether the quota criterion is satisfied.
A: 115; B: 675; C: 205 (modified divisor 9.78)
Determine the number of representatives for each state using Jefferson's method with a modified divisor of 9.78 . A: 11 B: 69 C. 20
State whether the quota criterion is satisfied. The quota criterion is \square because the number of seats assigned to at least one of the states \square the same as its \square quota either rounded up or down to the nearest integer

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Problem 1140

The city of Raleigh has 8300 registered voters. There are two candidates for city council in an upcoming election: Brown and Feliz. The day before the election, a telephone poll of 500 randomly selected registered voters was conducted. 176 said they'd vote for Brown, 294 said they'd vote for Feliz, and 30 were undecided.
Describe the population actually represented by this survey. All citizens of Raleigh All registered voters in Raleigh All registered voters with telephones in Raleigh The 500 voters surveyed The 176 voters who said they'd vote for Brown None of the above

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Problem 1141

h. Interpret the level of significance in the context of the study. - There is a 10%10 \% chance that there is a difference in the proportion of blonde and brunette college students who have a boyfriend. If the percent of all blonde college students who have a boyfriend is the same as the percent of all brunette college students who have a boyfriend and if another 638 blonde college students and 791 brunette college students are surveyed then there would be a 10%10 \% chance that we would end up falsely concuding that the population proportion of blonde college students who have a boyfriend is greater than the population proportion of brunette college students who have a boyfriend If the percent of all blonde college students who have a boyfriend is the same as the percent of all brunette college students who have a boyfriend and if another 638 blonde college students and 791 brunette college students are surveyed then there would be a 10%10 \% chance that we would end up falsely concuding that the proportion of these surveyed blonde and brunette college students who have a boyfriend differ from each other. There is a 10%10 \% chance that you will never get a boyfriend unless you dye your hair blonde.

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Problem 1142

Are job applicants with easy to pronounce last names less likely to get called for an interview than applicants with difficult to pronounce last names. 673 job applications were sent out with last names that are easy to pronounce and 761 identical job applications were sent out with names that were difficult to pronounce. 424 of the "applicants" with easy to pronounce names were called for an interview while 502 of the "applicants" with difficult to pronounce names were called for an interview. What can be concluded at the 0.10 level of significance? If the calculator asks, be sure to use the "Not Pooled" data option.
For this study, we should use z-test for the difference between two population proportions \square a. The null and alternative hypotheses would be: b. The test statistic \square zθ2=z \theta^{2}= (please show your answer to 3 decimal places.) c. The p -value == \square (Please show your answer to 4 decimal places.) d. The pp-value is \square α\alpha e. Based on this, we should \square fail to reject \checkmark \checkmark the null hypothesis.

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Problem 1143

Are job applicants with easy to pronounce last names less likely to get called for an interview than applicants with difficult to pronounce last names. 673 job applications were sent out with last names that are easy to pronounce and 761 identical job applications were sent out with names that were difficult to pronounce. 424 of the "applicants" with easy to pronounce names were called for an interview while 502 of the "applicants" with difficult to pronounce names were called for an interview. What can be concluded at the 0.10 level of significance? If the calculator asks, be sure to use the "Not Pooled" data option.
For this study, we should use \square zz-test for the difference between two population proportions a. The null and alternative hypotheses would be: H0 : = p2  -  (please enter a decimal) H1 :  p1  <  p2 \begin{array}{l} H_{0} \text { : } \\ = \\ \text { p2 } \\ \text { - }{ }^{\checkmark} \text { (please enter a decimal) } \\ H_{1} \text { : } \\ \text { p1 } \\ \text { < } \\ \text { p2 } \end{array} (Please enter a decimal) b. The test statistic \square \square (please show your answer to 3 decimal places.) c. The pp-value == 0.1209 \square (Please show your answer to 4 decimal places.)

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Problem 1144

Do men score higher on average compared to women on their statistics finals? Final exam scores of thirteen randomly selected male statistics students and twelve randomly selected female statistics students are shown below.
Male: 93776471726284708973638894\begin{array}{lllllllllllll}93 & 77 & 64 & 71 & 72 & 62 & 84 & 70 & 89 & 73 & 63 & 88 & 94\end{array}  Female: 834658814974627069665868\begin{array}{lllllllllllll}\text { Female: } 83 & 46 & 58 & 81 & 49 & 74 & 62 & 70 & 69 & 66 & 58 & 68\end{array} Assume both follow a Normal distribution. What can be concluded at the the α=0.01\alpha=0.01 level of significance level of significance?
For this study, we should use Select an answer a. The null and alternative hypotheses would be: H0H_{0} : Select an answer Select an answer Select an answer (6) (please enter a decimal) H1H_{1} : Select an answer Select an answer Select an answer (Please enter a decimal) b. The test statistic ? 0=0= \square (please show your answer to 3 decimal places.) c. The pp-value == \square (Please show your answer to 4 decimal places.) d. The pp-value is ? α\alpha e. Based on this, we should Select an answer the null hypothesis. f. Thus, the final conclusion is that ... The results are statistically insignificant at α=0.01\alpha=0.01, so there is statistically significant evidence to conclude that the population mean statistics final exam score for men is equal to the population mean statistics final exam score for women. The results are statistically significant at α=0.01\alpha=0.01, so there is sufficient evidence to conclude that the population mean statistics final exam score for men is more than the population mean statistics final exam score for women. The results are statistically significant at α=0.01\alpha=0.01, so there is sufficient evidence to conclude that the mean final exam score for the thirteen men that were observed is more than the mean final exam score for the twelve women that were observed. The results are statistically insignificant at α=0.01\alpha=0.01, so there is insufficient evidence to conclude that the population mean statistics final exam score for men is more than the population mean statistics final exam score for women.

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Problem 1145

Daily Caloric Intake According to one source, people eat on average 3630 calories per day. If the standard deviation is 325 calories, find the zz score for each raw score. Round zz scores to at least two decimal places.
Part: 0/3\mathbf{0} / \mathbf{3} \square
Part 1 of 3
The zz score corresponding to 2890 calories is \square .

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Problem 1146

(fofe)2fe=(Observed cell frequencyExpected cell frequency)2Expected cell frequency \frac{(f_o - f_e)^2}{f_e} = \frac{(\text{Observed cell frequency} - \text{Expected cell frequency})^2}{\text{Expected cell frequency}}

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Problem 1147

A biologist looked at the relationship between number of seeds a plant produces and the percent of those seeds that sprout. The results of the survey are shown below. \begin{tabular}{|c|r|r|r|r|r|l|l|l|} \hline Seeds Produced & 48 & 57 & 65 & 63 & 61 & 60 & 56 & 52 \\ \hline Sprout Percent & 57 & 54.5 & 50.5 & 50.5 & 53.5 & 51 & 61 & 66 \\ \hline \end{tabular} a. Find the correlation coefficient: r=r= \square Round to 2 decimal places. b. The null and alternative hypotheses for correlation are: H0:? ? =0H1:? ? 0\begin{array}{l} H_{0}: ? \quad \text { ? }=0 \\ H_{1}: ? \text { ? } \neq 0 \end{array}
The pp-value is: \square (Round to four decimal places) c. Use a level of significance of α=0.05\alpha=0.05 to state the conclusion of the hypothesis test in the context of the study. There is statistically insignificant evidence to conclude that there is a correlation between the number of seeds that a plant produces and the percent of the seeds that sprout. Thus, the use of the regression line is not appropriate. There is statistically significant evidence to conclude that a plant that produces more seeds will have seeds with a lower sprout rate than a plant that produces fewer seeds. There is statistically significant evidence to conclude that there is a correlation between the number of seeds that a plant produces and the percent of the seeds that sprout. Thus, the regression line is useful. There is statistically insignificant evidence to conclude that a plant that produces more seeds will have seeds with a lower sprout rate than a plant that produces fewer seeds. d. r2=r^{2}= \square (Round to two decimal places) e. Interpret r2r^{2} : 54%54 \% of all plants produce seeds whose chance of sprouting is the average chance of sprouting. Given any group of plants that all produce the same number of seeds, 54%54 \% of all of these plants will produce seeds with the same chance of sprouting. There is a 54%54 \% chance that the regression line will be a good predictor for the percent of seeds that sprout based on the number of seeds produced. There is a large variation in the percent of seeds that sprout, but if you only look at plants that produce a fixed number of seeds, this variation on average is reduced by 54%54 \%.

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Problem 1148

A biologist looked at the relationship between number of seeds a plant produces and the percent of those seeds that sprout. The results of the survey are shown below. \begin{tabular}{|c|l|r|r|r|r|l|l|l|} \hline Seeds Produced & 48 & 57 & 65 & 63 & 61 & 60 & 56 & 52 \\ \hline Sprout Percent & 57 & 54.5 & 50.5 & 50.5 & 53.5 & 51 & 61 & 66 \\ \hline \end{tabular} a. Find the correlation coefficient: r=r= \square 0.73-0.73 Round to 2 decimal places. b. The null and alternative hypotheses for correlation are: H0:ρΔ2=0H1:ρ20\begin{array}{l} H_{0}: \rho \Delta^{2}=0 \\ H_{1}: \rho \nabla^{2} \neq 0 \end{array}
The pp-value is: \square (Round to four decimal places) c. Use a level of significance of α=0.05\alpha=0.05 to state the conclusion of the hypothesis test in the context of the study. There is statistically insignificant evidence to conclude that there is a correlation between the number of seeds that a plant produces and the percent of the seeds that sprout. Thus, the use of the regression line is not appropriate. There is statistically significant evidence to conclude that a plant that produces more seeds will have seeds with a lower sprout rate than a plant that produces fewer seeds. There is statistically significant evidence to conclude that there is a correlation between the number of seeds that a plant produces and the percent of the seeds that sprout. Thus, the regression line is useful. There is statistically insignificant evidence to conclude that a plant that produces more seeds will have seeds with a lower sprout rate than a plant that produces fewer seeds. d. r2=r^{2}= \square 0.54 (Round to two decimal places)

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Problem 1149

Question 4
Adult men have heights with a mean of 69.0 inches and a standard deviation of 2.8 inches. Find the height of a man with a zz-score of 0.3571 (to 4 decimal places) \square

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Problem 1150

A newsgroup is interested in constructing a 90\% confidence interval for the proportion of all Americans who are in favor of a new Green initiative. Of the 546 randomly selected Americans surveyed, 403 were in favor of the initiative. Round answers to 4 decimal places where possible. a. With 90%90 \% confidence the proportion of all Americans who favor the new Green initiative is between \square and \square b. If many groups of 546 randomly selected Americans were surveyed, then a different confidence interval would be produced from each group. About \square percent of these confidence intervals will contain the true population proportion of Americans who favor the Green initiative and about \square percent will not contain the true population proportion.

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Problem 1151

Question 13
In a recent poll, 580 people were asked if they liked dogs, and 68\% said they did. Find the Margin of Error for this poll, at the 99%99 \% confidence level. Give your answer to four decimal places if possible. \square

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Problem 1152

A student was asked to find a 98%98 \% confidence interval for the population proportion of students who take notes using data from a random sample of size n=90n=90. Which of the following is a correct interpretation of the interval 0.1<p<0.310.1<p<0.31 ?
Check all that are correct. There is a 98\% chance that the proportion of the population is between 0.1 and 0.31 . The proportion of all students who take notes is between 0.1 and 0.31,98%0.31,98 \% of the time. With 98%98 \% confidence, a randomly selected student takes notes in a proportion of their classes that is between 0.1 and 0.31 . With 98%98 \% confidence, the proportion of all students who take notes is between 0.1 and 0.31 . There is a 98%98 \% chance that the proportion of notetakers in a sample of 90 students will be between 0.1 and 0.31 .

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Problem 1153

Aisha wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 77 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 7.1 and a standard deviation of 1.7. What is the 80%80 \% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Assume the data is from a normally distributed population. Round answers to 3 decimal places where possible. \square <μ<<\mu< \square

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Problem 1154

A hypothesis test produced a pp-value of 0.09981 . If the test was conducted at the significance level α=0.09\alpha=0.09, which of the following is correct about the result? Reject H0H_{0}; neither Type I error nor Type II error is possible. Do not reject H0H_{0}; a Type II error is possible. Reject H0H_{0}; a Type I error is possible. Reject H1H_{1}; a Type I error is possible. Do not reject H1H_{1}; a Type II error is possible.

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Problem 1155

A researcher wishes to examine the relationship between years of schooling completed and the number of pregnancies in young women. Her research discovers a linear relationship, and the least squares line is: y^=43x\hat{y}=4-3 x where x is the number of years of schooling completed and y is the number of pregnancies. The slope of the regression line can be interpreted in the following way: When amount of schooling increases by one year, the number of pregnancies tends to decrease by 3 . When amount of schooling increases by one year, the number of pregnancies tends to increase by 3 . When amount of schooling increases by one year, the number of pregnancies tends to increase by 4 . When amount of schooling increases by one year, the number of pregnancies tends to decrease by 4 .

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Problem 1156

myopenmath.com
The following table shows retail sales in drug stores in billions of dollars in the U.S. for years since 1995. \begin{tabular}{|c|c|} \hline Year & Retail Sales \\ \hline 0 & 85.851 \\ \hline 3 & 108.426 \\ \hline 6 & 141.781 \\ \hline 9 & 169.256 \\ \hline 12 & 202.297 \\ \hline 15 & 222.266 \\ \hline \end{tabular}
Let S(t)S(t) be the retails sales in billions of dollars in tt years since 1995. A linear model for the data is F(t)=9.44t+84.182F(t)=9.44 t+84.182.
Use the above scatter plot to decide whether the linear model fits the data well. The function is a good model for the data. The function is not a good model for the data Estimate the retails sales in the U. S. in 2013. \square billions of dollars. Use the model to predict the year that corresponds to retails sales of $237\$ 237 billion. \square

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Problem 1157

```latex \textbf{Problem:}
Given the following data on the average number of weekly hours worked by U.S. production workers from 1967 to 1996:
\begin{center} \begin{tabular}{|c|c|} \hline Year & Hours Worked \\ \hline 1967 & 38.0 \\ 1968 & 37.8 \\ 1969 & 37.7 \\ 1970 & 37.1 \\ 1971 & 36.9 \\ 1972 & 37.0 \\ 1973 & 36.9 \\ 1974 & 36.5 \\ 1975 & 36.1 \\ 1976 & 36.1 \\ 1977 & 36.0 \\ 1978 & 35.8 \\ 1979 & 35.7 \\ 1980 & 35.3 \\ 1981 & 35.2 \\ 1982 & 34.8 \\ 1983 & 35.0 \\ 1984 & 35.2 \\ 1985 & 34.9 \\ 1986 & 34.8 \\ 1987 & 34.8 \\ 1988 & 34.7 \\ 1989 & 34.6 \\ 1990 & 34.5 \\ 1991 & 34.3 \\ 1992 & 34.4 \\ 1993 & 34.5 \\ 1994 & 34.7 \\ 1995 & 34.5 \\ 1996 & 34.4 \\ \hline \end{tabular} \end{center}
1. Construct a scatter diagram and comment on the relationship, if any, between the variable Year and Hours Worked.
2. Determine and interpret the correlation for the year and hours worked. Based upon the value of the correlation, is your answer to the previous question reasonable?
3. Based upon the data given, estimate the average weekly hours worked this year. How confident are you in your estimate? You should use a linear regression model to make your prediction.
4. Assuming a linear correlation between these two variables, what will happen to the average weekly hours worked in the future? Is it possible for this pattern to continue indefinitely? Explain.

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Problem 1158

\begin{tabular}{|c|c|} \hline \multicolumn{2}{|c|}{\begin{tabular}{c} At a basketball game, you ask random students what \\ his or her favorite movie gener is to determine the \\ schools favorite type of movie. \end{tabular}} \\ \hline Population & * \\ \hline Sample & * \\ \hline What is biased? & * \\ \hline \begin{tabular}{c} How can it be \\ fixed? \end{tabular} & \\ \hline \end{tabular}

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Problem 1159

Math 142-8111 (FA'24) in Canvas andres salazar 12/09/24 3:54 AM Homework: HW \#12: Sections 9.1-9.3 Question 38, 9.3.6-T HW Score: 64.3%,32.1564.3 \%, 32.15 of 50 points Part 1 of 6 Points: 0 of 1 Save
Question list Listed below are the numbers of words spoken in a day by each member of eight different randomly selected couples. Complete parts (a) and (b) below. \begin{tabular}{lllcccccc} \hline Male & 15,710 & 27,243 & 1359 & 7676 & 19,266 & 15,734 & 13,532 & 25,101 \\ \hline Female & 24,464 & 13,027 & 18,834 & 17,762 & 12,841 & 16,316 & 15,904 & 19,201 \\ \hline \end{tabular} Question 37 Question 38 Question 39 Question 40 a. Use a 0.01 significance level to test the claim that among couples, males speak fewer words in a day than females.
In this example, μd\mu_{d} is the mean value of the differences dd for the population of all pairs of data, where each individual difference dd is defined as the words spoken by the male minus words spoken by the female. What are the null and alternative hypotheses for the hypothesis test? H0=μd\mathrm{H}_{0}=\mu_{\mathrm{d}} \square word(s) H1=μd\mathrm{H}_{1}=\mu_{\mathrm{d}} \square \square word(s) (Type integers or decimals. Do not round.) Question 41 Question 42

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Problem 1160

Homework: HW \#12: Sections 9.1-9.3 Question 40, 9.3.11-T HW Score: 71.53%,35.7671.53 \%, 35.76 of 50 points Part 1 of 6 Points: 0 of 1 Save
Question list Question 40 Question 41 */, Question 42 x/s Question 43 Question 44 Question 45
The following data lists the ages of a random selection of actresses when they won an award in the category of Best Actress, along with the ages of actors when they won in the category of Best Actor. The ages are matched according to the year that the awards were presented. Complete parts (a) and (b) below. \begin{tabular}{lllllllllll} \hline Actress (years) & 25 & 25 & 33 & 26 & 39 & 28 & 24 & 41 & 33 & 37 \\ \hline Actor (years) & 60 & 40 & 33 & 38 & 31 & 33 & 49 & 42 & 34 & 43 \\ \hline \end{tabular} a. Use the sample data with a 0.05 significance level to test the claim that for the population of ages of Best Actresses and Best Actors, the differences have a mean less than 0 (indicating that the Best Actresses are generally younger than Best Actors). In this example, μd\mu_{\mathrm{d}} is the mean value of the differences d for the population of all pairs of data, where each individual difference d is defined as the actress's age minus the actor's age. What are the null and altermative hypotheses for the hypothesis test? H0:μd\mathrm{H}_{0}: \mu_{\mathrm{d}} \square year(s) H1:μd\mathrm{H}_{1}: \mu_{\mathrm{d}} \square years) (Type integers or decimals. Do not round.)
Help me solve this View an example Get more help - Clear all check answer

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Problem 1161

Suppose XN(2,6)X \sim N(2, 6). What value of xx has a zz-score of three?

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Problem 1162

For a population with μ=100\mu = 100 and σ=20\sigma = 20, what is the XX value corresponding to z=0.25z = 0.25? 90 110 105 95 Need Help? Read It

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Problem 1163

H0:μ=20H_0: \mu = 20 Ha:μ<20H_a: \mu < 20 Use the following information: n=48n = 48, X=18\overline{X} = 18, and σ=4.6\sigma = 4.6 To find the test statistic (Step 2).

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Problem 1164

4) In a pilot study we collected data on how many words 5 participants can remember. The data for this sample is 4, 5, 6, 7, and 8. In the general population, people can remember 8 words. What is the t value for this sample data? QUESTION 5

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Problem 1165

Confidence in banks: A poll conducted asked a random sample of 1268 adults in the United States how much confidence they had in banks and other financial institutions. A total of 144 adults said that they had a great deal of confidence. An economist claims that less than 11% of U.S. adults have a great deal of confidence in banks. Can you conclude that the economist's claim is true? Use both α=0.05\alpha = 0.05 and α=0.10\alpha = 0.10 levels of significance and the PP-value method with the TI-84 Plus calculator.
Part 1 of 5
(a) State the appropriate null and alternate hypotheses.
H0:p=0.11H_0: p = 0.11 H1:p<0.11H_1: p < 0.11
This hypothesis test is a left-tailed test.
Part: 1/5
Part 2 of 5
(b) Compute the value of the test statistic. Round the answer to at least two decimal places. z=z =

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Problem 1166

1\checkmark 1 2\checkmark 2 3\checkmark 3 4\checkmark 4 5\checkmark 5 6 ×7\times 7 8 9 10
Caffeine in Tea Bags In a recent study, researchers found that the average amount of caffeine in tea bags is 26 mg . The standard deviation of the population is 3 mg . If a random sample of 46 newly manufactured tea bags has a mean of 25.1 mg of caffeine, can it be concluded that the new brand of tea bags contain less caffeine than the average found in the recent study? Use α=0.05\alpha=0.05, and use the PP-value method with tables.
Part: 0/50 / 5
Part 1 of 5 (a) State the hypotheses and identify the claim. H0: (Choose one) H1: claim  not claim \begin{array}{l} H_{0}: \square \text { (Choose one) } \nabla \\ H_{1}: \square \text { claim } \\ \text { not claim } \end{array}

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Problem 1167

Question 7 of 16 (1 point) Question Attempt: 1 of 1 Time Remaining tarks "general anxiety" of an individual, with higher GAS scores corresponding to more anxiety. (Dr. Elbod's assessment of anxiety is based on a variety of measurements, both physiological and psychological.) y^=8.320.27x\hat{y}=8.32-0.27 x. This line, along with a scatter plot of the sample data, is shown below. \begin{tabular}{|c|c|} \hline GAS score, x\boldsymbol{x} & \begin{tabular}{c} Sleep time, y\boldsymbol{y} \\ (in hours) \end{tabular} \\ \hline 6.5 & 6.1 \\ \hline 7.0 & 6.5 \\ \hline 5.9 & 7.9 \\ \hline 1.0 & 6.9 \\ \hline 1.6 & 8.6 \\ \hline 4.0 & 6.6 \\ \hline 3.0 & 7.1 \\ \hline 3.6 & 8.2 \\ \hline 9.1 & 6.1 \\ \hline 3.5 & 7.5 \\ \hline 7.9 & 6.9 \\ \hline 2.0 & 8.3 \\ \hline 9.2 & 5.3 \\ \hline 8.1 & 5.4 \\ \hline 5.2 & 6.2 \\ \hline \end{tabular} Send data to calculator Send data to Excel
Based on the study's data and the regression line, answer the following. (a) From the regression equation, what is the predicted sleep time (in hours) when the GAS score is 8.1 ? Round your answer to one or more decimal places.

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Problem 1168

A political newspaper, Keeping It Political, is interested in the distances voters in Vanville County live from their nearest polling station. Keeping It Political has hired you to estimate the population mean of all such distances for voters in this county. To estimate this population mean, you select a random sample of 23 Vanville County voters and record the distance (in km ) each voter lives from their nearest polling station. Assume the population is approximately normally distributed. Based on your sample, follow the steps below to construct a 95%95 \% confidence interval for the population mean of all the distances voters in Vanville County live from their nearest polling station. (If necessary, consult a list of formulas.) (a) Click on "Take Sample" to see the results for your random sample. \begin{tabular}{|c|c|c|} \hline Number of voters & Sample mean & \begin{tabular}{c} Sample standard \\ deviation \end{tabular} \\ \hline 23 & 4.192 & 2.041 \\ \hline \end{tabular}
Enter the values of the sample size, the point estimate of the mean, the sample standard deviation, and the critical value you need for your 95%95 \% confidence interval. (Choose the correct critical value from the table of critical values provided.) When you are done, select "Compute". \begin{tabular}{|l|l|} \hline Sample size: \\ \square \end{tabular} \begin{tabular}{|c|l|} \hline \begin{tabular}{c} Confidence \\ level \end{tabular} & Critical value \\ \hline 99%99 \% & t0.005=2.819t_{0.005}=2.819 \\ \hline 95%95 \% & t0.025=2.074t_{0.025}=2.074 \\ \hline 90%90 \% & t0.050=1.717t_{0.050}=1.717 \\ \hline \end{tabular}

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Problem 1169

Explain the differences between the chi-square test for independence and the chi-square test for homogeneity. What are the similarities?
The difference is that the chi-square test for independence compares two characteristics from one population and the chi-square test for homogeneity compares one characteristic from more than one population.
What are the similarities? Select all that apply.
A. The hypotheses are the same. B. The procedures are the same. C. The assumptions are the same. D. The data are the same.

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Problem 1170

In a study, researchers wanted to measure the effect of alcohol on the hippocampal region, the portion of the brain responsible for long-term memory storage, in adolescents. The researchers randomly selected 11 adolescents with alcohol use disorders to determine whether the hippocampal volumes in the alcoholic adolescents were less than the normal volume of 9.02 cm39.02 \mathrm{~cm}^{3}. An analysis of the sample data revealed that the hippocampal volume is approximately normal with no outliers and x=8.18 cm3\overline{\mathrm{x}}=8.18 \mathrm{~cm}^{3} and s=0.7 cm3\mathrm{s}=0.7 \mathrm{~cm}^{3}. Conduct the appropriate test at the α=0.01\alpha=0.01 level of significance.
State the null and alternative hypotheses. H0:μ=9.02H1:μ<9.02\begin{array}{l} H_{0}: \mu=9.02 \\ H_{1}: \mu<9.02 \end{array} (Type integers or decimals. Do not round.) Identify the tt-statistic. t0= (R t_{0}=\square \text { (R }

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Problem 1171

Eat your cereal: Boxes of cereal are labeled as containing 14 ounces. Following are the weights, in ounces, of a sample of 12 boxes. It is reasonable to assume that the population is approximately normal. 13.04 14.01 13.13 13.14 13.12 13.08 13.21 14.01 13.11 13.10 13.17 13.18
Send data to Excel
Part: 0 / 2
Part 1 of 2
(a) Construct a 98% confidence interval for the mean weight. Round the answers to at least three decimal places.
A 98% confidence interval for the mean weight is <μ< \Box < \mu < \Box .

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Problem 1172

Homework 93 Question 2 of 10 (1 point) I Question Attempt 1 of 3 Chikaodili
Tax-Exempt Properties A tax collector wishes to see if the mean values of the tax-exempt properties are different for two cities. The values of the tax-exempt properties for the two samples are shown. The data are given in millions of dollars. At α=0.05\alpha=0.05, is there enough evidence to support the tax collector's claim Español that the means are different? Assume the variables are normally distributed and the variances are unequal. Use a 718371-83 Plus/TI-84 Plus calculator. \begin{tabular}{cccc|cccc} \multicolumn{5}{c|}{ City A } & \multicolumn{3}{c}{ City B } \\ \hline 22 & 14 & 8 & 25 & 82 & 72 & 5 & 4 \\ 2 & 30 & 44 & 11 & 75 & 12 & 9 & 12 \\ 19 & 7 & 5 & & 68 & 81 & 2 & 60 \end{tabular} Send data to Excel
Use μ1\mu_{1} for the mean value of tax-exempt properties in City A .
Part 1 of 4 (a) State the hvootheses and identify the claim with the correct hypothesis. H0H_{0} : Try again thyslaim Save For Later Submit Assignment Skip Part Recheck - 2024 McGraw Hill LLC. All Rights Reserved. Terms of Use Privacy Center Accessibility 4:52 PM 12/9/2024

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Problem 1173

Demand for consumer goods increases as the unemployment rate decreases. \square Studies find that consumption of vitamin C reduces the number and severity of colds that people get.
The price of bread and canola oil both increase sharply after the prairies experience a drought during the growing season.
1. cause-and-effect relationship
2. common-cause factor
3. reverse cause-and-effect relationship
4. accidental relationship
5. presumed relationship

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Problem 1174

Q: Recording to the table below: (5 Points) \begin{tabular}{|c|c|c|c|c|c|c|c|} \hline & Entity 1 & Entity 2 & Entity 3 & Entity 4 & Entity 5 & Entity 6 & Entity 7 \\ \hline Arrival time per hr & 0 & 1 & 2 & 8 & 11 & 18 & Entity 723 \\ \hline Service per hr & 4 & 2 & 5 & 4 & 1 & 5 & 4 \\ \hline \end{tabular}
1. Manual Simulation; if we simulate the following system for 25 hrs only. What is the start time in system for entity number 5? (1 Point) a) 15 b) 39 c) 10 d) 12
2. What is the utilization of the system during the 25 hrs simulation? (1 Point) a) 90%90 \% b) 94%94 \% c) 96%96 \% d) 98%98 \%
3. What is the time-average number of entities in queue during the 25 hrs ? (1 Point) a) 3 b) 2.2 c) 2.5 d) 2.8\underline{2.8}
4. What are the entities-average number of entities in queue during the 25 hrs? (1 Point) a) 2\underline{2} b) 2.9 c) 4.0 d) 1.9
5. What is the total number of entities in queue during the 25 hrs? (1 Point) a) 30 b) 50\underline{50} c) 45

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Problem 1175

A pediatrician wants to determine the relation that may exist between a child's height and head circumference. She randomly selects 5 children and measures their height and head circumference. The data are summarized below. Complete parts (a) through (f) below. Height (inches), xx 27.75 27 25.5 26.5 26.75 Head Circumference (inches), yy 17.6 17.5 17.1 17.3 17.3
H0:β1=0H_0: \beta_1 = 0 H1:β10H_1: \beta_1 \ne 0
H0:β0=0H_0: \beta_0 = 0 H1:β00H_1: \beta_0 \ne 0
Determine the P-value for this hypothesis test. P-value =0.008= 0.008 (Round to three decimal places as needed.) What is the conclusion that can be drawn?
A. Do not reject H0H_0 and conclude that a linear relation does not exist between a child's height and head circumference at the level of significance α=0.01\alpha = 0.01. B. Reject H0H_0 and conclude that a linear relation does not exist between a child's height and head circumference at the level of significance α=0.01\alpha = 0.01. C. Reject H0H_0 and conclude that a linear relation exists between a child's height and head circumference at the level of significance α=0.01\alpha = 0.01. D. Do not reject H0H_0 and conclude that a linear relation exists between a child's height and head circumference at the level of significance α=0.01\alpha = 0.01.
(e) Use technology to construct a 95% confidence interval about the slope of the true least-squares regression line. Lower bound: Upper bound: (Round to three decimal places as needed.)

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Problem 1176

A regression line to predict number of calories consumed in a day based on number of grams of fat consumed in a day is Calories^=400+18(Fat) \hat{Calories} = 400 + 18 \cdot (Fat) . We can interpret the slope as:
If a person consumes 1 gram of fat more, predicted number of calories goes up by 400 .. If amount of calories consumed is 18 more, predicted amount of fat goes up by 1 gram. If amount of calories consumed is 1 more, predicted amount of fat goes up by 18 grams. If amount of fat consumed is 18 hour more per week, predicted pulse rate goes up 1.. If amount of fat consumed is 1 gram more, predicted number of calories goes up by 18.. If a person consumes 1 calorie more, predicted amount of fat goes up by 400 grams.

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Problem 1177

18 Maya is conducting a study on the impact of a new math tutoring program in her classes. Which method would introduce the least bias into her sample? A choosing the students with the highest grades from each class to use in her sample (B) randomly selecting students from each of her classes to participate in the study (C) selecting only students from her afternoon classes, to be part of the sample (D) selecting only female students from all her math classes to use in her sample

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Problem 1178

Government funding: The following table presents the budget (in millions of dollars) for selected organizations that received U.S. government funding for arts and culture in both 2006 and last year.
Organization | 2006 | Last Year --- | --- | --- Organization 1 | 460 | 440 Organization 2 | 247 | 227 Organization 3 | 142 | 161 Organization 4 | 124 | 156 Organization 5 | 95 | 166 Organization 6 | 18 | 45 Organization 7 | 2 | 3
Part: 0 / 3 Part 1 of 3 Compute the least-squares regression line for predicting last year's budget from the 2006 budget Round the slope and yy-intercept to four decimal places as needed. The equation for the least-squares regression line is y^=\hat{y} =.

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Problem 1179

SAT scores: Assume that in a given year the mean mathematics SAT score was 605605, and the standard deviation was 136136. A sample of 7676 scores is chosen. Use the TI-84 Plus calculator.
Part 1 of 5
(a) What is the probability that the sample mean score is less than 589589? Round the answer to at least four decimal places.
The probability that the sample mean score is less than 589589 is 0.15150.1515.
Part 2 of 5
(b) What is the probability that the sample mean score is between 575575 and 610610? Round the answer to at least four decimal places.
The probability that the sample mean score is between 575575 and 610610 is ____.

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Problem 1180

Eat your cereal: Boxes of cereal are labeled as containing 14 ounces. Following are the weights, in ounces, of a sample of 16 boxes. It is reasonable to assume that the population is approximately normal. 14.07 13.98 14.16 14.17 14.11 14.03 14.16 13.98 14.06 14.05 14.12 14.13 14.19 14.17 14.05 14.04
Part: 0/2
Part 1 of 2
(a) Construct a 95% confidence interval for the mean weight. Round the answers to at least three decimal places.
A 95% confidence interval for the mean weight is <μ<<\mu<.

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Problem 1181

Treating circulatory disease: Angioplasty is a medical procedure in which an obstructed blood vessel is widened. In some cases, a wire mesh tube, called a stent, is placed in the vessel to help it remain open. A study was conducted to compare the effectiveness of a bare metal stent with one that has been coated with a drug designed to prevent reblocking of the vessel. A total of 5330 patients received bare metal stents, and of these, 844 needed treatment for reblocking within a year. A total of 1122 received drug-coated stents, and 148 of them required treatment within a year. Can you conclude that the proportion of patients who needed retreatment differs between those who received bare metal stents and those who received drug-coated stents? Let p1p_1 denote the proportion of patients with bare metal stents who needed retreatment and p2p_2 denote the proportion of patients with drug-coated stents who needed retreatment. Use the α=0.05\alpha = 0.05 level of significance and the PP-value method with the TI-84 Plus calculator.
Part: 0 / 4
Part 1 of 4
State the appropriate null and alternate hypotheses. H0H_0: H1H_1:
This hypothesis test is a (Choose one) test.

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Problem 1182

In order to answer the question below, which of the following types of study would be the most appropriate? What percentage of the population reads a newspaper on a regular basis?
A. Case-control study B. Experiment without blinding C. Experiment with double blinding D. Observational study E. Experiment with single blinding

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Problem 1183

The police department in a large city has 177 new officers to be apportioned among six high-crime precincts. Crimes by precinct are shown in the following table. Use Adams's method with d=16d = 16 to apportion the new officers among the precincts.
Precinct | A | B | C | D | E | F ------- | -------- | -------- | -------- | -------- | -------- | -------- Crimes | 450 | 529 | 801 | 235 | 321 | 425
Precinct | A ------- | -------- Number of Apportioned Police Officers | 29 (Type an integer.)
Precinct | B ------- | -------- Number of Apportioned Police Officers | (Type an integer.)

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Problem 1184

Listed below are the lead concentrations (in μ\mug/g) measured in different Ayurveda medicines. Ayurveda is a traditional medical system commonly used in India. The lead concentrations listed here are from medicines manufactured in the United States. Assume that a simple random sample has been selected. Use a 0.05 significance level to test the claim that the mean lead concentration for all such medicines is less than 14.0 μ\mug/g.
3.01 6.51 5.95 5.47 20.51 7.51 12.03 20.49 11.47 17.50
Identify the null and alternative hypotheses.
H0H_0: μ=14.0\mu = 14.0 H1H_1: μ<14.0\mu < 14.0
(Type integers or decimals. Do not round.)
Identify the test statistic.
(Round to two decimal places as needed.)

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Problem 1185

ion 10.2 Homework Question 8, 10.2.21-T HW Score: 69.48%,15.2969.48 \%, 15.29 of 22 points Part 1 of 5 Points: 0.29 of 1 Save
Twenty years ago, 53%53 \% of parents of children in high school felt it was a serious problem that high school students were not being taught enough math and science. A recent survey found that 216 of 700 parents of children in high school felt it was a serious problem that high school students were not being taught enough math and science. Do parents feel differently today than they did twenty years ago? Use the α=0.1\alpha=0.1 level of significance.
Because np0(1p0)=n p_{0}\left(1-p_{0}\right)= \square \square 10, the sample size is \square 5%5 \% of the population size, and the sample \square the requirements for testing the hypothesis \square satisfied. (Round to one decimal place as needed.)

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Problem 1186

The null and alternative hypotheses are given. Determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed. What parameter is being tested? H0:σ=110H_0: \sigma = 110 H1:σ>110H_1: \sigma > 110
Is the hypothesis test left-tailed, right-tailed, or two-tailed? Right-tailed test Two-tailed test Left-tailed test
What parameter is being tested? Population proportion Population standard deviation Population mean

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Problem 1187

Homework: HW \#14: Sections 11.1-11.2 Question 21, 11.2.11-T HW Score: 54.52%,16.3654.52 \%, 16.36 of 30 points Points: 0 of 1 Save
Question list Question 21 Question 22 Question 23 Question 24 x/s Question 25 Question 26
The accompanying table shows results of challenged referee calls in a major tennis tournament. Use a 0.05 significance level to test the claim that the gender of the tennis player is independent of whether a call is overturned.
Click the icon to view the table. A. H0\mathrm{H}_{0} : The gender of the tennis player is not independent of whether a call is overturned. H1\mathrm{H}_{1} : The gender of the tennis player is independent of whether a call is overturned. H0\mathrm{H}_{0} : Male tennis players are more successful in overturning calls than female players. H1\mathrm{H}_{1} : Male tennis players are not more successful in overturning calls than female players. . H0\mathrm{H}_{0} : Male tennis players are not more successful in overturning calls than female players H1\mathrm{H}_{1} : Male tennis players are more successful in overturning calls than female players. H0\mathrm{H}_{0} : The gender of the tennis player is independent of whether a call is overturned. H1\mathrm{H}_{1} : The gender of the tennis player is not independent of whether a call is overturned.
Determine the test statistic χ2=\chi^{2}= \square (Round to three decimal places as needed.) Print Done \begin{tabular}{|l|c|c} \hline & \multicolumn{2}{|c}{ Was the Challenge to the Call Successful? } \\ \hline & Yes & No \\ \hline Men & 343 & 719 \\ \hline Women & 462 & 825 \\ \hline \end{tabular} \qquad 111\sqrt[1]{11} 7 (1,1) Clear all Check answer View an example Get more help -

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Problem 1188

abes warming. The respondents who answered yes when asked if there is solid evidence that the earth is getting warmer were then independent of the choice for the
Question 23 Question 24 Question 25 Question 26 x/5x / 5 Question 27 Question 28 \begin{tabular}{l|ccc} & Human activity & Natural patterns & Don't know \\ \hline Male & 344 & 140 & 40 \\ Female & 333 & 166 & 37 \end{tabular}
Click here to view the chi-square distribution table. \qquad \qquad Identify the null and alternative hypotheses. H0\mathrm{H}_{0} : \square d \square H1\mathrm{H}_{1} : \square \square

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Problem 1189

andres salazar 12/09/24 7:45 PM Homework: HW \#14: Sections 11.1-11.2 Question 7, 11.1.13-T HW Score: 60.28%,18.0860.28 \%, 18.08 of 30 points Part 1 of 5 Points: 0 of 1 Save
Question list x/s Question 4 */, Question 5 *// Question 6 * Question 7 x// Question 8 *// Question 9 View an example
Get more help - Clear all check answer

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Problem 1190

Question list x/s Question 4 x/s Question 5 ×// Question 6 * Question 7 x/s Question 8 x/s Question 9
The table below lists the frequency of wins for different post positions in a horse race. A post position of 1 is closest to the inside rall, so that horse has the shortest distance to run. (Because the number of horses varies from year to year, only the first 10 post positions are included.) Use a 0.05 significance level to test the claim that the likelihood of winning is the same for the different post positions. Based on the result, should bettors consider the post position of a horse race? Post Position Wins 1 20 2 13 \begin{tabular}{c} 3 \\ 12 \\ \hline \end{tabular} 4 16 5 13 6 8 7 8 8 12 9 5 10 11 Save
Determine the null and alternative hypotheses H0\mathrm{H}_{0} : Wins occur with equal frequency in the different post positions. H1H_{1} : At least one post position has a different frequency of wins than the others. Calculate the test statistic, x2x^{2} x2=x^{2}= \square (Round to three decimal places as needed.) Clear all check answer Help me solve this View an example Get more help - Dec 9 7:47 US

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Problem 1191

A random sample of 49 measurements from one population had a sample mean of 14, with sample standard deviation 3. An independent random sample of 64 measurements from a second population had a sample mean of 16, with sample standard deviation 4. Test the claim that the population means are different. Use level of significance 0.01.
USE SALT
(a) What distribution does the sample test statistic follow? Explain.
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations. The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations. The Student's t. We assume that both population distributions are approximately normal with known standard deviations. The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
(b) State the hypotheses.
H0:μ1=μ2H_0: \mu_1 = \mu_2; H1:μ1>μ2H_1: \mu_1 > \mu_2 H0:μ1=μ2H_0: \mu_1 = \mu_2; H1:μ1μ2H_1: \mu_1 \neq \mu_2 H0:μ1μ2H_0: \mu_1 \neq \mu_2; H1:μ1=μ2H_1: \mu_1 = \mu_2 H0:μ1=μ2H_0: \mu_1 = \mu_2; H1:μ1<μ2H_1: \mu_1 < \mu_2

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Problem 1192

Click here to view the standard normal distribution table (page 1) Click here to view the standard normal distribution table (page 2). (a) What is the probability that a randomly selected time interval between eruptions is longer than 94 minutes?
The probability that a randomly selected time interval is longer than 94 minutes is approximately .3372 . (Round to four decimal places as needed.) (b) What is the probability that a random sample of 14 time intervals between eruptions has a mean longer than 94 minutes?
The probability that the mean of a random sample of 14 time intervals is more than 94 minutes is approximately \square. (Round to four decimal places as needed.)

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Problem 1193

10.3 Homework Question 2, 10.3.5 Part 4 of 6 HW Score: 25.6%,5.3825.6 \%, 5.38 of 21 points Points: 0.38 of 1 Save
To test H0:μ=100H_{0}: \mu=100 versus H1:μ100H_{1}: \mu \neq 100, a simple random sample of size n=19n=19 is obtained from a population that is known to be normally distributed. Answer parts (a)-(e). Click here to view the t-Distribution Area in Right Tail. t=2.432\mathrm{t}=2.432 (Round to three decimal places as needed) (b) If the researcher decides to test this hypothesis at the α=0.01\alpha=0.01 level of significance, determine the critical values.
The critical value(s) is/are 2.878,2.878-2.878,2.878. (Use a comma to separate answers as needed. Round to three decimal places as needed.) (c) Draw a t-distribution that depicts the critical region(s). Which of the following graphs shows the critical region(s) in the t-distribution? A. B. c. (d) Will the researcher reject the null hypothesis? A. The researcher will reject the null hypothesis since the test statistic is not in the rejection region. B. The researcher will reject the null hypothesis since the test statistic is in the rejection region. C. There is not sufficient evidence for the researcher to reject the null hypothesis since the test statistic is not in the rejection region. D. There is not sufficient evidence for the researcheirto reject the null hypothesis since the test statistic is in the rejection region. Clear all Check answer

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Problem 1194

38. Joan's Nursery specializes in custom-designed landscaping for residential areas. The estimated labor cost associated with a particular landscaping proposal is based on the number of plantings of trees, shrubs, and so on to be used for the project. For cost-estimating purposes, managers use two hours of labor time for planting a medium-sized tree. A random sample of 30 plantings during the past month was selected. Based on the sample, the mean number of planting time is 2.2 hours with a standard deviation of 0.516. With a 0.05 level of significance, use the PMACC process to test to see whether the population mean treeplanting time differs from two MoM_o hours.

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Problem 1195

10.3 Homework Question 15, 10.3.36-T HW Score: 28.57\%, 6 of 21 points Part 13 of 13 Points: 0 of 1 Save
Suppose the mean IQ score of people in a certain country is 103. Suppose the director of a college obtains a simple random sample of 37 students from that country and finds the mean IQ is 107 with a standard deviation of 13.8. Complete parts (a) through (d) below.
Find the test statistic for this hypothesis test. 0.88 (Round to two decimal places as needed.)
Find the PP-value for this hypothesis test. 0.192 (Round to three decimal places as needed.)
Write a conclusion for the test. Choose the correct answer below. A. Do not reject H0\mathrm{H}_{0}. There is not sufficient evidence to conclude that the mean IQ score of people in the country is greater than 105 at the α=0.01\alpha=0.01 level of significance. B. Reject H0\mathrm{H}_{0}. There is not sufficient evidence to conclude that the mean IQ score of people in the country is greater than 105 at the α=0.01\alpha=0.01 level of significance. C. Reject H0\mathrm{H}_{0} - There is sufficient evidence to conclude that the mean IQ score of people in the country is greater than 105 at the α=0.01\alpha=0.01 level of significance. D. Do not reject H0H_{0}. There is sufficient evidence to conclude that the mean IQ score of people in the country is greater than 105 at the α=0.01\alpha=0.01 level of significance. (d) Based on the results of parts (a)-(c), write a few sentences that explain the difference between "accepting" the statement in the null hypothesis versus "not rejecting" the statement in the null hypothesis.
One should \square rather than \square the null hypothesis. If one \square the null hypothesis, this indicates that the \square is a specific value, such as 103, 104, or 105, and so the same data have been used to conclude that the \square is three different values. If one \square the null hypothesis, this indicates that the \square could be 103, 104, or 105 or even some other value; we are simply not ruling them out as the value of the \square Therefore, \square the null hypothesis can lead to contradictory conclusions, whereas \square does not. an example Get more help - Clear all Check answer

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Problem 1196

Determine μxˉ\mu_{\bar{x}} and σxˉ\sigma_{\bar{x}} from the given parameters of the population and sample size. μ=76,σ=7,n=49μxˉ=\begin{array}{l} \mu=76, \sigma=7, n=49 \\ \mu_{\bar{x}}=\square \end{array} \square

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Problem 1197

Question 8 (1 point) What is the value of aa used in describing the confidence interval shown below.
98% μ\mu a 0.01 b 0.02 c 0.04 d 0.05

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Problem 1198

A survey was given to a random sample of 850 voters in the United States to ask about their preference for a presidential candidate. Of those surveyed, 459 respondents said that they preferred Candidate A. Determine a 95%95 \% confidence interval for the proportion of people who prefer Candidate A, rounding values to the nearest thousandth.

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Problem 1199

A study of a local high school tried to determine the mean number of text messages that each student sent per day. The study surveyed a random sample of 128 students in the high school and found a mean of 197 messages sent per day with a standard deviation of 88 messages. At the 95%95 \% confidence level, find the margin of error for the mean, rounding to the nearest whole number. (Do not write ±\pm ).

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Problem 1200

8. (2 punti) Il livello di significatività α\alpha del test è A) la probabilità di rifiutare correttamente l'ipotesi corretta B) uguale ad 1 meno la probabilità di commettere un errore di I tipo C) la probabilità di rifiutare l'ipotesi nulla quando l'ipotesi nulla è corretta D) uguale alla probabilità di commettere un errore di II tipo E) nessuna delle precedenti

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