Statistical Inference

Problem 301

Shown to the right is a certain population, in billions, for seven selected years from 1950 through 2006. Using a graphing utility's logistic regression option, we obtain the logistic growth model shown below for population, $\mathrm{f}(\mathrm{x})$ , in billions, x years after 1949. How well does the function model the data for 2006? $f(x)=\frac{11.82}{1+3.81 e^{-0.027 x}}$ \begin{tabular}{|c|c|} \hline X, Number of Years after 1949 & y, Population (billions) \\ \hline $1(1950)$ & 2.6 \\ \hline $11(1960)$ & 3.0 \\ \hline $21(1970)$ & 3.7 \\ \hline $31(1980)$ & 4.5 \\ \hline $41(1990)$ & 5.3 \\ \hline $51(2000)$ & 6.1 \\ \hline $57(2006)$ & 6.5 \\ \hline \end{tabular}
For 2006 , the function $\square$ the population to one decimal place. accurately predictś slightly underestimates slightly overestimates

See Solution

Problem 302

You are given the sample mean and the population standard deviation. Use this information to construct the $90 \%$ and $95 \%$ confidence intervals for the population mean. Interpret esults and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals.
A random sample of 35 home theater systems has a mean price of $\$ 120.00$ . Assume the population standard deviation is $\$ 19.60$ .
Construct a $90 \%$ confidence interval for the population mean. he $90 \%$ confidence interval is $\square$ $\square$ Round to two decimal places as needed.)

See Solution

Problem 303

The use of the formula for margin of error requires a large sample. For each of the following combinations of $n$ and $\hat{p}$ , indicate whether the sample size is large enough for use of this formula to be appropriate. (a) $n=100$ and $\hat{p}=0.60$ The sample size is large enough. The sample size is not large enough. (b) $n=40$ and $\hat{p}=0.25$ The sample size is large enough. The sample size is not large enough. (c) $n=50$ and $\hat{p}=0.25$ The sample size is large enough. The sample size is not large enough. (d) $n=80$ and $\hat{p}=0.10$ The sample size is large enough. The sample size is not large enough.

See Solution

Problem 304

If all other quantities remain the same, how does the indicated change affect the minimum sample size requirement? (a) Increase in the level of confidence (b) Increase in the error tolerance (c) Increase in the population standard deviation (a) How does an increase in the level of confidence affect the minimum sample size requirement? Choose the correct answer below.
An increase in the level of confidence $\square$ the minimum sample size required.

See Solution

Problem 305

The table to the right shows the results of a survey in which 2579 adults from Country A, 1129 adults from Country B, and 1059 adults from Country C were asked if human activity contributes to global warming. Complete parts (a), (b), and (c).
Adults who say that human activity contributes to global warming \begin{tabular}{|l|r|} \hline Country A & $66 \%$ \\ \hline Country B & $85 \%$ \\ \hline Country C & $92 \%$ \\ \hline \end{tabular} (a) Construct a $90 \%$ confidence interval for the proportion of adults from Country A who say human activity contributes to global warming. ( $0.645,0.675$ ) (Round to three decimal places as needed.) (b) Construct a $90 \%$ confidence interval for the proportion of adults from Country B who say human activity contributes to global warming. $(0.833,0.867)$ (Round to three decimal places as needed.) (c) Construct a $90 \%$ confidence interval for the proportion of adults from Country C who say human activity contributes to global warming. ( $\square$ (Round to three decimal places as needed.)

See Solution

Problem 306

n randomized, double-blind clinical trials of a new vaccine, monkeys were randomly divided into two groups. Subjects in group 1 received the new vaccine while subjects in group 2 received a control vaccine. After the second dose, 123 of 662 subjects in the experimental group (group 1) experienced drowsiness as a side effect. After the second dose, 80 of 543 of the subjects in the control group (group 2) experienced drowsiness as a side effect. Does the evidence suggest that a higher proportion of subjects in group 1 experienced drowsiness as a side effect than subjects in group 2 at the $\alpha=0.10$ level of significance? F. The data come from a population that is normally distributed.
Determine the null and alternative hypotheses. $\begin{array}{l} H_{0}: p_{1}=p_{2} \\ H_{1}: p_{1}>p_{2} \end{array}$
Find the test statistic for this hypothesis test. 1.78 (Round to two decimal places as needed.)
Determine the P-value for this hypothesis test. $\square$ (Round to three decimal places as needed.)

See Solution

Problem 307

10.77 An experiment is conducted to determine if the use of a special chemical additive with a standard fertilizer accelerates plant growth. Ten locations are included in the study. At each location, two plants growing in close proximity are treated. One is given the standard fertilizer, the other the standard fertilizer with the chemical additive. Plant growth after four weeks is measured in centimeters. Do the following data substantiate the claim that use of the chemical additive accelerates plant growth? State the assumptions that you make and devise an appropriate test of the hypothesis. Take $\alpha=.05$ . \begin{tabular}{l|cccccccccc} \hline & \multicolumn{8}{|c}{ Location } \\ \cline { 2 - 10 } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline \begin{tabular}{c} Without \\ additive \\ \begin{tabular}{c} With \\ additive \end{tabular} \end{tabular} 20 & 23 & 17 & 22 & 19 & 32 & 25 & 18 & 21 & 19 \\ \hline \end{tabular}

See Solution

Problem 308

Your answer is partially correct.
Is a t-Distribution Appropriate? A sample with size $n=12$ has $\bar{x}=7.6$ and $s=1.6$ . The dotplot for this sample is given below.
Indicate whether or not it is appropriate to use the $t$ -distribution.
Yes $\leqslant$
If it is appropriate, give the degrees of freedom for the $t$ -distribution and give the estimated standard error. If it is not appropriate, enter -1 in both of the answer fields below.
Enter the exact answer for the degrees of freedom and round your answer for the standard error to two decimal places. $d f=$ i $\square$ standard error = $\square$ eTextbook and Media Hint Attempts: 1 of 5 used Submit Answer

See Solution

Problem 309

View Policies
Current Attempt in Progress Is a t-Distribution Appropriate? A sample with size $n=75$ has $\bar{x}=18.92$ , and $s=10.1$ . The dotplot for this sample is given below.
Indicate whether or not it is appropriate to use the $t$ -distribution. $\square$ If it is appropriate, give the degrees of freedom for the $t$ -distribution and give the estimated standard error. If it is not appropriate, enter -1 in both of the answer fields below.
Enter the exact answer for the degrees of freedom and round your answer for the standard error to two decimal places. $d f=$ $\square$ i standard error = $\square$ i eTextbook and Media Save for Later Attempts: 0 of 5 used Submit Answer

See Solution

Problem 310

How Many Birds Do Domestic Cats Kill? "Domestic cats kill many more wild birds in the United States than scientists thought," states a recent article. ${ }^{1}$ Researchers used a sample of $n=140$ households in the US with cats to estimate that $35 \%$ of household cats in the US hunt outdoors. A separate study ${ }^{2}$ used KittyCams to record all activity of $n=55$ domestic cats that hunt outdoors. The video footage showed that the mean number of kills per week for these cats was 2.4 with a standard deviation of 1.51 . Find a $99 \%$ confidence interval for the mean number of kills per week by US household cats that hunt outdoors.
Round your answers to three decimal places.
The 99 confidence interval is i $\square$ to i $\square$ ${ }^{1}$ Milius, S., "Cats kill more than one billion birds each year," Science News, 183(4), February 23, 2013, revised March 8, 2014. Data approximated from information give in the article. ${ }^{2}$ Loyd KAT, et al., "Quantifying free-roaming domestic cat predation using animal-borne video cameras," Biological Conservation, 160(2013), 183-189.

See Solution

Problem 311

According to one source, $52 \%$ of plane crashes are due at least in part to pilot error. Suppose that in a random sample of 100 separate airplane accidents, 67 of them were due at least in part to pilot error. Complete parts (a) and (b) below. $\begin{array}{l} H_{0}: p=0.52 \\ H_{a}: p \neq 0.52 \end{array}$ (Type integers or decimals. Do not round.) Determine the test statistic. $z=3.00$ (Round to two decimal places as needed.) Determine the p -value. $p \text {-value }=0.002$ (Round to three decimal places as needed.) What is the proper conclusion? Reject $\mathrm{H}_{0}$ . There is sufficient evidence to conclude that the population proportion is not 0.52 . (Type an integer or a decimal. Do not round.) b. Determine the correct interpretaction of the rejection decision.
The percentage of plane crashes due to pilot error $\square$ significantly different from $52 \%$ because the p-value is $\square$ the significance level.

See Solution

Problem 312

Pepcid A study of 74 patients with ulcers was conducted in which they were prescribed 40 mg of Pepcid ${ }^{\mathrm{TM}}$ . After 8 weeks, 58 reported ulcer healing. (a) Obtain a point estimate for the proportion of patients with ulcers receiving Pepcid who will report ulcer healing. (b) Verify that the requirements for constructing a confidence interval about $p$ are satisfied. (c) Construct a $99 \%$ confidence interval for the proportion of patients with ulcers receiving Pepcid who will report ulcer healing. (d) Interpret the confidence interval.

See Solution

Problem 313

A 2018 poll of 2236 randomly selected U.S. adults found that $36.58 \%$ planned to watch at least a "fair amount" of a particular sporting event in 2018. In 2014, 46\% of U.S. adults reported planning to watch at least a "fair amount." a. Does this sample give evidence that the proportion of U.S. adults who planned to watch the 2018 sporting event was less than the proportion who planned to do so in 2014? Use a 0.05 significance level. b. After conducting the hypothesis test, a further question one might ask is what proportion of all U.S. adults planned to watch at least a "fair amount" of the sporting event in 2018. Use the sample data to construct a $90 \%$ confidence interval for the population proportion. How does your confidence interval support your hypothesis test conclusion? a. State the null and alternative hypotheses. Let p be the proportion of U.S. adults that planned to watch at least a "fair amount" of the sporting event. $\begin{array}{l} H_{0}: p=0.46 \\ H_{a}: p<0.46 \end{array}$ (Type integers or decimals. Do not round.) Compute the z-test statistic. $z=-8.97$ (Round to two decimal places as needed.) Compute the p-value. $\mathrm{p} \text {-value }=$ $\square$ (Round to three decimal places as needed.)

See Solution

Problem 314

Let's go back to the television question: According to the Buckert and Howell research firm the time spent by kids watching television per year can be modelled by a normal distribution with a mean of 1200 hours and a standard deviation of 180 hours.
What percent of children watch between 1000 and 1500 hours of television? Shade the distribution belon
1. The average reading score for BSHS students is 580 with a standard deviation of 50 points. What percent of students are: a) Between 500 and 600? d) More than 680 ? b) Between 650 and 750 ? c) Less than 450 ?

See Solution

Problem 315

Weekly Hours and Grade Average \begin{tabular}{|c|c|} \hline Hours Worked & Overall Grade Average \\ \hline 15 & 86 \\ \hline 30 & 72 \\ \hline 27 & 77 \\ \hline 25 & 83 \\ \hline 16 & 87 \\ \hline 20 & 90 \\ \hline 12 & 94 \\ \hline \end{tabular}
Based on the correlation coefficient for the data, what type of linear association exists between hours worked and overall grade average? (A) Strong negative (B) Weak nenative

See Solution

Problem 316

1. (8 marks) Brock University is studying student midterm grades in a particular Canadian geography course. They randomly select 8 students from the course and document their scores from the midterm, which are given below (in percentages): $81,72,79,48,61,68,84,91$
Assuming the population is approximately normal and using a 0.10 significance level, test the claim that the mean test score for the entire class is greater than 70. a) State the null and alternative hypotheses. b) Calculate the test statistic (round it to 3 dechmal places). c) Find the critical value (round it to 3 decimal places). d) Make a decision and write a conclusion based on your answers from parts b) and c).

See Solution

Problem 317

MORE BENEFITS OF EATING ORGANIC
Using specific data, we find a significant difference in the proportion of fruit flies surviving after 13 days between those eating organic potatoes and those eating conventional (not organic) potatoes. This exercise asks you to conduct a hypothesis test using additional data. In this case, we are testing $\begin{array}{l} H_{0}: p_{o}=p_{c} \\ H_{a}: p_{o}>p_{c} \end{array}$ where $p_{o}$ and $p_{c}$ represent the proportion of fruit flies alive at the end of the given time frame of those eating organic food and those eating conventional food, respectively. Use a $5 \%$ significance level.
Effect of Organic Soybeans After 5 Days After 5 days, the proportion of fruit flies eating organic soybeans still alive is 0.89 , while the proportion still alive eating conventional soybeans is 0.85 . The standard error for the difference in proportions is 0.022 .
What is the value of the test statistic? Round your answer to two decimal places. $z=$
What is the $p$ -value? Round your answer to three decimal places. $p$ -value $=$ $\square$ What is the conclusion? $\square$ Is there evidence of a difference? $\square$

See Solution

Problem 318

Find the critical tscore to three decimal places when the Confidence Level is $93 \%$ and the sample size is 32 A. 2.245 B. 2.891 C. 1.682 D. 1.877

See Solution

Problem 319

A data set includes data from student evaluations of courses. The summary statistics are $n=95, \bar{x}=4.07, s=0.53$ . Use a 0.05 significance level to test the claim that the population of student course evaluations has a mean equal to 4.25. Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.
What are the null and alternative hypotheses? A. $H_{0}: \mu=4.25$ B. $H_{0}: \mu \neq 4.25$ $\mathrm{H}_{1}: \mu<4.25$ $H_{1}: \mu=4.25$ C. $H_{0}: \mu=4.25$ D. $H_{0}: \mu=4.25$ $\mathrm{H}_{1}: \mu>4.25$ $\mathrm{H}_{1}: \mu \neq 4.25$
Determine the test statistic. $\square$ (Round to two decimal places as needed.) Determine the P -value. (Round to three decimal places as needed.)

See Solution

Problem 320

It is known that roughly 2/3 of all human beings have a dominant right foot or eye. Is there also right-sided dominance in kissing behavior? An article reported that in a random sample of 115 kissing couples, both people in 73 of the couples tended to lean more to the right than to the left. (Use a = 0.05.) USE SALT (a) If 2/3 of all kissing couples exhibit this right-leaning behavior, what is the probability that the number in a sample of 115 who do so differs from the expected value by at least as much as what was actually observed? (Round your answer to four decimal places.) (b) Does the result of the experiment suggest that the 2/3 figure is implausible for kissing behavior? State the appropriate null and alternative hypotheses. Ho: P = 2/3 Hp < 2/3 Ho P = 2/3 H: p + 2/3 O Ho: p = 2/3 HP ≤ 2/3 OHP = 2/3 H:p> 2/3 Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) == P-value =

See Solution

Problem 321

A firm sent out a random sample of 500 shipments in a new type of heavy-duty packaging. Eighteen of them damaged upon arrival at their destinations. Is it appropriate to use the methods of this section to construct a confidence interval for the proportion of shipments in heavy-duty packaging that are damaged upon arrival? If not, explain why not. Yes No, because a sample of size 500 is not large enough. No, because either $n \hat{p}$ or $n(1-\hat{p})$ is less than 10 .

See Solution

Problem 322

Suppose the mean wait-time for a telephone reservation agent at a large airline is 45 seconds. A manager with the airline is concerned that business may be lost due to customers having to wait too long for an agent. To address this concern, the manager develops new airline reservation policies that are intended to reduce the amount of time an agent needs to spend with each customer. A random sample of 250 customers results in a sample mean wait-time of 44.4 seconds with a standard deviation of 4.3 seconds. Using $\alpha=0.05$ level of significance, do you believe the new policies were effective in reducing wait time? Do you think the results have any practical significance?
Determine the null and alternative hypotheses. $\begin{array}{l} \mathrm{H}_{0}: \mu=45 \text { seconds } \\ \mathrm{H}_{1}: \mu<45 \text { seconds } \end{array}$
Calculate the test statistic. $t_{0}=-2.21$ (Round to two decimal places as needed.) Calculate the P -value. P -value $=0.014$ (Round to three decimal places as needed.) State the conclusion for the test. A. Reject $\mathrm{H}_{0}$ because the P -value is less than the $\alpha=0.05$ level of significance. B. Reject $\mathrm{H}_{0}$ because the $P$ -value is greater than the $\alpha=0.05$ level of significance. C. Do not reject $H_{0}$ because the $P$ -value is less than the $\alpha=0.05$ level of significance. D. Do not reject $H_{0}$ because the $P$ -value is greater than the $\alpha=0.05$ level of significance.
State the conclusion in context of the problem. There $\square$ sufficient evidence at the $\alpha=0.05$ level of significance to conclude that the new policies were effective. is not is Clear all Final check

See Solution

Problem 323

Listen
Provide the null and alternative hypotheses to test whether $\mu$ is less than 50. a) $H_{0}: \mu \geq 50$ vs. $H_{a}: \mu<50$ b) $H_{0}: \mu \leq 50$ vs. $H_{a}: \mu>50$

See Solution

Problem 324

Find the value of $Z_{\text {crit }}$ and state the conclusion for $H_{0}: \mu=\mu_{0}$ vs. $H_{a}$ : $\mu \neq \mu_{0}, \alpha=0.10, Z_{\text {deta }}=1.6$ . a) $\pm 1.28$ , reject $H_{0}$ . b) $\pm 2.575$ , do not reject $H_{0}$ .

See Solution

Problem 325

Determine the parameter to be tested, and the test to be performed.
Part: 0 / 6 $\square$
Part 1 of 6
A simple random sample of size 65 has mean $\bar{x}=7.26$ and the standard deviation is $s=1.9$ . Can you conclude that the population mean is greater than 9 ? Determine the parameter to be tested.
The parameter to be tested is the population (Choose one) $\square$

See Solution

Problem 326

7) Express the original claim in symbolic form. 긴
Claim: $20 \%$ of adults smoke A) $0=0.2$ B) $u \neq 0.2$ C) $0=0.2$ D) $u=0.2$
A formal hypothesis test is to be conducted using the claim that the mean body temperature is equal to $98.6^{\circ} \mathrm{F}$ . What is the null hypothesis and how is it denoted? B) $\mathrm{H}_{O}: \mu=98.6^{\circ} \mathrm{F}$ A) $\mathrm{H}_{\mathrm{O}}: p=98.6^{\circ} \mathrm{F}$ D) $\mathrm{H}_{\mathrm{O}}: \mu \neq 98.6^{\circ} \mathrm{F}$

See Solution

Problem 327

26) Formulate the indicated conclusion in nontechnical terms. Be sure to address the original claim. Carter Motor Company claims that its new sedan, the Libra, will average better than 32 miles 26) $\qquad$ per gallon in the city. Assuming that a hypothesis test of the claim has been conducted and that the conclusion is to reject the null hypothesis, state the conclusion in nontechnical terms. A) There is sufficient evidence to support the claim that the mean is less than 32 miles per gallon. B) There is not sufficient evidence to support the claim that the mean is less than 32 miles per gallon. C) There is sufficient evidence to support the claim that the mean is greater than 32 miles per gallon. D) There is not sufficient evidence to support the claim that the mean is greater than 32 miles pergallon. 27) Use the given information to find the $P$ -value. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null 27) $\qquad$ hypothesis). With $H_{1}: p \neq 3 / 5$ , the test statistic is $z=0.78$ . A) 0.2177 ; reject the null hypothesis B) 0.4354 ; reject the null hypothesis

See Solution

Problem 328

A smart phone manufacturer is interested in constructing a 90\% confidence interval for the proportion of smart phones that break before the warranty expires. 92 of the 1683 randomly selected smart phones broke before the warranty expired. Round answers to 4 decimal places where possible. a. With $90 \%$ confidence the proportion of all smart phones that break before the warranty expires is between $\square$ and $\square$ b. If many groups of 1683 randomly selected smart phones are selected, then a different confidence interval would be produced for each group. About $\square$ percent of these confidence intervals will contain the true population proportion of all smart phones that break before the warranty expires and about $\square$ percent will not contain the true population proportion.

See Solution

Problem 329

C8 Q2 V1: A random sample of 16 students is taken from a larger normal population of students in a multi-lecture course and the sample mean (average) of their final grades is found to be 62.25. Suppose it is known that the sample standard deviation is 3.2. You wish to create a $92 \%$ confidence interval for the true population mean. Which of the following statements is most correct? a. The $\mathrm{ta} / 2, \mathrm{n}-1$ (found from R) used to calculate the $92 \%$ confidence interval is 0.04 . b. The $\mathrm{ta} / 2, \mathrm{n}-1$ (found from R) used to calculate the $92 \%$ confidence interval is 2.24854 . c. The $t a / 2, \mathrm{n}-1$ (found from R ) used to calculate the $92 \%$ confidence interval is 1.478367 . d. The $t a / 2, \mathrm{n}-1$ (found from R ) that is needed to calculate the $92 \%$ confidence interval is 1.877739 .

See Solution

Problem 330

C8 Q5 V3: A random sample of 100 students is taken from a larger population of students in a multi-lecture course and the sample mean (average) of their final grades is found to be 63.25. Suppose it is known that the population standard deviation is 3.4. You wish to determine whether the average grade for the population differs from 62.75 . Use R to get the needed $Z$ critical value and then create a $99 \%$ Z confidence interval for the average final grade for the population by hand. Choose the correct statement below. a. The test is significant at the $1 \%$ significance level because 62.75 is inside the confidence interval you created. b. The test is not significant at the $1 \%$ significance level because 63.25 is inside the confidence interval you created. c. The test is not significant at the $1 \%$ significance level because 62.75 is inside the confidence interval you created. d. The test is significant at the $1 \%$ significance level because 62.75 is outside the confidence interval you created.

See Solution

Problem 331

C8 Q13 V1: The Excel file STATISTICSSTUDENTSSURVEYFORR contains the column BEFPULSEMIN (a numerical variable that measures student pulses before completing an online survey). For education purposes, consider this dataset to be a sample of size 60 taken from a much wider population for statistics students. Use R to find a $94 \%$ confidence interval for the mean of the population. Choose the most correct (closest) statement below. a. Your confidence interval is $(72.88831,75.91169)$ beats per minute and is narrower than a $90 \%$ confidence interval. b. Your confidence interval is $(72.88831,75.91169)$ beats per minute and it is wider than a $90 \%$ confidence interval created from this data. c. Your confidence interval is $(72.95127,75.84873)$ beats per minute and it is narrower than a $90 \%$ confidence interval created from this data. d. Your confidence interval is $(72.95127,75.84873)$ beats per minute and it is wider than a $90 \%$ confidence interval created from this data.

See Solution

Problem 332

C9 Q2 V3 The Excel file STATISTICSSTUDENTSSURVEYFORR contains the column BEFPULSEMIN (a numerical variable that measures student pulses before completing an online survey). For education purposes, consider this dataset to be a sample of size 60 taken from a much larger population for statistics students. You test, with a significance level of 0.002 , whether there is significant evidence that the average beats per minute for the population exceeds 71.75 beats per minute. Choose the most correct statement. a. You test Ho: $\mu<71.75$ versus Ha : $\mu>71.75$ and you obtain a p-value of 0.0004356 and the result is not significant b. You test Ho: $\mu>=71.75$ versus Ha: $\mu<71.75$ and you obtain a p-value of 0.9996 and the result is not significant c. You test Ho: $\mu<=71.75$ versus Ha: $\mu>71.75$ and you obtain a pvalue of 0.0004356 and the result is significant d. You test Ho: $\mu<=71.75$ versus Ha: $\mu>71.75$ and you obtain a p-value of 0.0008713 and the result is not significant

See Solution

Problem 333

C9 Q5 V3: A random sample of 100 students is taken from a larger population of students in a multi-lecture course and the sample mean (average) of their final grades is 66. Suppose it is known that the population standard deviation is 7. You test whether there is significant evidence that the population average differs from 67 , using a level of significance of $4 \%$ . Choose the most correct statement. Note: You will have to calculate the test statistic by hand, but you can use R to get your pvalue. a. Your pvalue statement is $2 \mathrm{P}(\mathrm{Z}>1.43)$ , your pvalue is 0.1531 , and your decision is to reject Ho. b. Your pvalue statement is $2 P(Z>1.43)$ , your pvalue is 0.1531 and your decision is to fail to reject Ho c. Your pvalue statement is $2 P(Z>1.43)$ , your pvalue is 0.3062 and your decision is to fail to reject Ho. d. Your pvalue statement is $2 \mathrm{P}(\mathrm{Z}>1.43)$ , your pvalue is 0.0766 and your decision is reject Ho.

See Solution

Problem 334

C9 Q6 V1: The Excel file STATISTICSSTUDENTSSURVEYFORR contains the column BEFPULSEMIN (a numerical variable that measures student pulses before completing an online survey). For education purposes, consider this dataset to be a sample of size 60 taken from a much larger population for statistics students. You test whether there is significant evidence that the average beats per minute for the population differs from 76 beats per minute, using a level of significance of $3 \%$ . Choose the most correct statement. a. Your pvalue statement is $2 \mathrm{P}(\mathrm{t}>2.1179)$ , your pvalue is 0.019205 , and your decision is reject Ho b. Your pvalue statement is $2 \mathrm{P}(\mathrm{t}>2.1179)$ , your pvalue is 0.9808 , and your decision is to fail to reject Ho c. Your pvalue statement is $2 \mathrm{P}(\mathrm{t}>2.1179)$ , your pvalue is 0.03841 , and your decision is to fail to reject Ho d. Your pvalue statement is $2 \mathrm{P}(\mathrm{t}>2.1179)$ , your pvalue is 0.07682 , and your decision is to fail to reject Ho

See Solution

Problem 335

Company records indicate that less than $24 \%$ of employees commute to work using public transportation. For a sample of 207 employees, you find that $18 \%$ of them take public transportation to work. Test the claim using a $1 \%$ level of significance. a. What type of test will be used in this problem?
Select an answer b. What evidence justifies the use of this test? Check all that apply $n p \geq 5$ and $n q \geq 5$ The sample standard deviation is not known The original population is approximately normal The population standard deviation is not known The population standard deviation is known The sample size is larger than 30 There are two different samples being compared c. Enter the null hypothesis for this test. $H_{0}$ : $\square$ ? $\square$ d. Enter the alternative hypothesis for this test. $H_{1}$ : $\square$ ? $\square$

See Solution

Problem 336

6. A medical researcher claims that $5 \%$ of children under 18 years of age have asthma. In a random sample of 250 children under 18 years of age, $9.6 \%$ say they have asthma. At $\alpha=0.01$ , is there enough evidence to reject the researcher's claim?

See Solution

Problem 337

A doctor is concerned that nausea may be a side effect of Tamiflu, but is not certain because nausea is common for people who have the flu. She researched some past studies, and found that about $36 \%$ of people who get the flu and do not use Tamilflu experience nausea. She then collected data on 2992 patients who were taking Tamiflu, and found that 1017 experienced nausea. Use a 0.01 significance level to test the claim that the percentage of people who take Tamiflu for the relief of flu symtoms and experience nausea is $36 \%$ . a. What type of test will be used in this problem?
Select an answer b. What evidence justifies the use of this test? Check all that apply
The sample standard deviation is not known $n p>5$ and $n q>5$ The sample size is larger than 30
There are two different samples being compared The original population is approximately normal The population standard deviation is not known The population standard deviation is known c. Enter the null hypothesis for this test. $H_{0}$ $\square$ ? $\square$ d. Enter the alternative hypothesis for this test. $H_{1}: ? \sim ? \sim$ $\square$

See Solution

Problem 338

(d) Level 99\%, sample size 29
2. A sample of size $n=35$ has sample mean $\bar{x}=34.85$ and sample standard deviation $s=17.9$ . (a) Construct a $98 \%$ confidence interval for the population mean $\mu$ . (b) If the confidence level were $95 \%$ , would the confidence interval be narrower or wider than the interval constructed in part (a)?

See Solution

Problem 339

Complete the sentence below. The $\qquad$ denoted $\hat{p}$ , is given by the formula $\hat{p}=$ $\qquad$ , where $x$ is the number of individuals with a specified characteristic in a sample of n individuals.
The $\square$ $\square$ , denoted $\hat{p}$ , is given by the formula $\hat{p}=$ $\square$ , where x is the number of individuals with a specified characteristic in a sample of n individuals.

See Solution

Problem 340

A farmer is concerned that a change in fertilizer to an organic variant might change his crop yield. He subdivides six lots and uses the old fertilizer on one half of each lot and the new fertilizer on the other half. The accompanying data file shows the crop yields with the old and new fertilizers.
Click here for the Excel Data File Let the difference be defined as Old - New. a. Specify the competing hypotheses that determine whether there is any difference between the average crop yields from the use of the different fertilizers. \begin{tabular}{|l|l|l|l|l|l|} \hline $H_{0}:$ & $\mu_{D}$ & $=$ & versus $H_{A}:$ & $\mu_{D}$ & $\neq$ \\ \hline \end{tabular} \begin{tabular}{|l|l|l|l|} \hline Lot & Old & New & \\ \hline 1 & 11 & 12 & \\ \hline 2 & 10 & 9 & \\ \hline 3 & 9 & 12 & \\ \hline 4 & 10 & 10 & \\ \hline 5 & 12 & 10 & \\ \hline 6 & 11 & 12 & \\ \hline & & & \\ \hline \end{tabular} b. Assuming that differences in crop yields are normally distributed, calculate the value of the test statistic.
Note: Negative value should be indicated by a minus sign. Round final answer to 3 decimal places.
Test statistic $\square$

See Solution

Problem 341

An experiment was conducted to determine whether giving candy to dining parties resulted in greater tips. The mean tip percentages and standard deviations are given in the accompanying table along with the sample sizes. Assume that the two samples are independent simple random samples selected from normally distributed populations, and \begin{tabular}{|c|c|c|c|c|} \hline & $\mu$ & $n$ & $\bar{x}$ & $\mathbf{s}$ \\ \hline No candy & $\mu_{1}$ & 36 & 19.34 & 1.38 \\ \hline Two candies & $\mu_{2}$ & 36 & 21.74 & 2.42 \\ \hline \end{tabular} do not assume that the population standard deviations are equal. Complete parts (a) and (b). a. Use a 0.05 significance level to test the claim that giving candy does result in greater tips.
What are the null and alternative hypotheses? A. $H_{0}: \mu_{1}=\mu_{2}$ B. $H_{0}: \mu_{1}=\mu_{2}$ $\mathrm{H}_{1}: \mu_{1}<\mu_{2}$ $H_{1}: \mu_{1}>\mu_{2}$ C. $\begin{array}{l} H_{0}: \mu_{1}=\mu_{2} \\ H_{1}: \mu_{1} \neq \mu_{2} \end{array}$ D. $H_{0}: \mu_{1} \neq \mu_{2}$ $\mathrm{H}_{1} \cdot \mu_{1} \mathrm{H}_{2}$
The test statistic, t , is $\square$ (Round to two decimal places as needed.)

See Solution

Problem 342

Question 16, 7.2.9 HW Score: 62.5\%, 15 of 24 Part 1 of 2 points Points: 0 of 1 Save
A data set includes 105 body temperatures of healthy adult humans having a mean of $98.7^{\circ} \mathrm{F}$ and a standard deviation of $0.64^{\circ} \mathrm{F}$ . Construct a $99 \%$ confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of $98.6^{\circ} \mathrm{F}$ as the mean body temperature? Click here to view at 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 $\mu$ ? $\square$ ${ }^{\circ} \mathrm{F}<\mu<$ $\square$ ${ }^{\circ} \mathrm{F}$ (Round to three decimal places as needed.) n example Get more help - Clear all Check answer

See Solution

Problem 343

You receive a brochure from a large university. The brochure indicates that the mean class size for full-time faculty is fewer than 32 students. You want to test this claim. You randomly select 18 classes taught by full-time faculty and determine the class size of each. The results are shown in the table below. At $\alpha=0.10$ , can you support the university's claim? Complete parts (a) through (d) below. Assume the population is normally distributed.
37 31 30 31 35 33 41 24 22 29 뭄 29 33 28 27 (a) Write the claim mathematically and identify $\mathrm{H}_{0}$ and $\mathrm{H}_{\mathrm{a}}$ .
Which of the following correctly states $\mathrm{H}_{0}$ and $\mathrm{H}_{\mathrm{a}}$ ? A. $\mathrm{H}_{0}: \mu>32$ B. $H_{0}: \mu \geq 32$ C. $\mathrm{H}_{0}: \mu=32$ $H_{a}: \mu \leq 32$ $H_{a}: \mu<32$ D. $H_{0}: \mu<32$ E. $H_{0}: \mu \leq 32$ $H_{a}: \mu \geq 32$ $H_{a}: \mu>32$ F. $\mathrm{H}_{0}: \mu=32$ $H_{a}: \mu \neq 32$ (b) Use technology to find the P -value. $P=$ $\square$ (Round to three decimal places as needed.)

See Solution

Problem 344

A car company says that the mean gas mileage for its luxury sedan is at least 22 miles per gallon (mpg). You believe the claim is incorrect and find that a random sample of 8 cars has a mean gas mileage of 21 mpg and a standard deviation of 2 mpg . At $\alpha=0.025$ , test the company's claim. Assume the population is normally distributed. Click here to view the t-distribution table. Click here to view page 1 of the normal table. Click here to view page 2 of the normal table.
Which sampling distribution should be used and why? A. Use a t-sampling distribution because the B. Use a normal sampling distribution because $\mathrm{n}>30$ . C. Use a t-sampling distribution because the D. Use a t-sampling distribution because $\mathrm{n}<30$ . population is normal, and $\sigma$ is known. E. Use a normal sampling distribution because the population is normal, and $\sigma$ is known. F. Use a normal sampling distribution because the population is normal, and $\sigma$ is unknown.

See Solution

Problem 345

A car company says that the mean gas mileage for its luxury sedan is at least 22 miles per gallon ( mpg ). You believe the claim is incorrect and find that a random sample of 8 cars has a mean gas mileage of 21 mpg and a standard deviation of 2 mpg . At $\alpha=0.025$ , test the company's claim. Assume the population is normally distributed. Click here to view the t-distribution table. Click here to view page 1 of the normal table. Click here to view page 2 of the normal table.
Which sampling distribution should be used and why? A. Use a t-sampling distribution because the B. Use a normal sampling distribution because $n>30$ . population is normal, and $\sigma$ is unknown. C. Use a t-sampling distribution because the D. Use a t-sampling distribution because $\mathrm{n}<30$ . population is normal, and $\sigma$ is known. E. Use a normal sampling distribution because the F. Use a normal sampling distribution because the population is normal, and $\sigma$ is known. population is normal, and $\sigma$ is unknown.
State the appropriate hypotheses to test. A. $H_{0}: \mu \neq 22$ B. $H_{0}: \mu \geq 22$ $H_{a}: \mu=22$ $H_{a}: \mu<22$ C. $H_{0}: \mu=22$ D. $H_{0}: \mu \leq 22$ $H_{a}: \mu \neq 22$ $H_{a}: \mu>22$

See Solution

Problem 346

ence Question 21, 7.2.27-T Part 1 of 2 HW Score: 90.97\%, 21.83 of 24 points Points: 0.5 of 1 Save:
Assume that all grade-point averages are to be standardized on a scale between 0 and 4 . How many grade-point averages must be obtained so that the sample mean is within 0.014 of the population mean? Assume that a $98 \%$ confidence level is desired. If using the range rule of thumb, $\sigma$ can be estimated as $\frac{\text { range }}{4}=\frac{4-0}{4}=1$ . Does the sample size seem practical?
The required sample size is $\square$ (Round up to the nearest whow number as needed.)

See Solution

Problem 347

isted below are systolic blood pressure measurements ( mm Hg ) taken from the right and left arms of the same woman. Assume imple random sample and that the differences have a distribution that is approximately normal. Use a 0.10 significance level to te neasurements from the two arms. What can be concluded? \begin{tabular}{llllll} Right arm & $=$ & 145 & 139 & 127 & 134 \\ Left arm \end{tabular} $=$ \begin{tabular}{ll} 175 & 160 \\ 185 & 140 \end{tabular} A. $H_{0}: \mu_{d}=0$ B. $H_{0}: \mu_{d} \neq 0$ $\mathrm{H}_{1}: \mu_{\mathrm{d}}<0$ $\mathrm{H}_{1}: \mu_{d}>0$ C. $H_{0}: \mu_{d} \neq 0$ D. $\begin{array}{l} H_{0}: \mu_{d}=0 \\ H_{1}: \mu_{d} \neq 0 \end{array}$
Identify the test statistic. $\mathrm{t}=-2.92$ (Round to two decimal places as needed.) Identify the P -value. P -value $=$ $\square$ (Round to three decimal places as needed.) example Get more help - Clear all

See Solution

Problem 348

interested to know if age is a factor in whether women wouts buy a particular garment depending on its quality. A researcher samples 3 age groups and cach mana is asked to rate the garment as excellent, average or poor. Test the iypothesis, at a 5\% level significance, that rating is not related to age group.
Rating Age group \begin{tabular}{|l|l|l|l|} \hline & $15-20$ & $21-30$ & $31-60$ \\ \hline Excellent & 40 & 47 & 46 \\ \hline Average & 51 & 74 & 57 \\ \hline Poor & 29 & 19 & 37 \\ \hline \end{tabular}

See Solution

Problem 349

The Affordable Care Act (ACA) signed into law by President Barrack Obama in 2010 was the most significant regulatory reform of the United States healthcare system since the passage of Medicare and Medicaid in 1965. The law was designed to expand insurance coverage and reduce the costs of health care. Some elements of the law such are banning insurance companies from denying coverage to people with pre-existing conditions are popular while other provisions such as the individual mandate are unpopular. Opinions on the law vary with some wanting to keep the law as it it, others favor fixing the law and some favor repealing the law in its entirity. The middle position has been gaining traction recently and in a Rasmussen Reports poll conducted January 5, 2017, 542 out of 967 randomly selected likely voters want Congress and the President to fix the law a piece at a time. a. Find the point estimate for the proportion of Americans who want Congress and the President to fix the law a piece at a time. Round your answer to 4 decimal places. $\square$ b. Find the margin of error for a $95 \%$ confidence interval. Round your answer to 4 decimal places. $\square$ c. Construct the $95 \%$ confidence interval for the proportion of Americans who want Congress and the President to fix the law a piece at a time. Enter your answer as an open interval of the form $(a, b)$ and round to 3 decimal places. $\square$ d. Can you conclude that more than half of Americans want Congress and the President to fix the law a piece at a time? Yes, the entire confidence is above 0.50 . No conclusions can be drawn since the confidence interval contains 0.50 . No, the entire confidence interval is below 0.50 .

See Solution

Problem 350

On average, a banana will last 6 days from the time it is purchased in the store to the time it is to eat. Is the mean time to spoil different if the banana is hung from the ceiling? The data shor an experiment with 14 bananas that are hung from the ceiling. Assume that that distribution of population is normal. $6.5,6.1,7.6,6.7,5,7.7,8.1,7.6,8.6,8.3,5.3,7.5,6.4,8.3$
What can be concluded at the the $\alpha=0.05$ level of significance level of significance? a. For this study, we should use
Select an answer b. The null and alternative hypotheses would be: $H_{0}$ : $\square$ Select an answer $\square$ $H_{1}$ : $\square$ Select an answer $\square$ c. The test statistic ? = $\square$ (please show your answer to 3 decimal places.) d. The $p$ -value $=$ $\square$ (Please show your answer to 4 decimal places.) e. The $p$ -value is $\square$ $\alpha$ f. Based on this, we should Seltht an answer $\qquad$ g. Thus, the final conclusion is that ... The data suggest that the population mean time that it takes for bananas to spoil if $t$ hung from the ceiling is not significantly different from 6 at $\alpha=0.05$ , so there is stat insignificant evidence to conclude that the population mean time that it takes for bai spoil if they are hung from the ceiling is different from 6. The data suggest the populaton mean is significantly different from 6 at $\alpha=0.05$ , so statistically significant evidence to conclude that the population mean time that it ta bananas to spoil if they are hung from the ceiling is different from 6. The data suggest the population mean is not significantly different from 6 at $\alpha=0.0$ \$ there is statistically insignificant evidence to conclude that the population mean time takes for bananas to spoil if they are hung from the ceiling is equal to 6 .

See Solution

Problem 351

At a PGA tournament (the Honda Classics at Palm Beach Gardens, Florida) the following scores were posted for eight randomly selected golfers for two consecutive days. At $\alpha=0.10$ , is there evidence of a difference in mean scores for the two days. \begin{tabular}{|l|cccccccc|c|} \hline Golfer & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline Thursday & 67 & 65 & 68 & 68 & 68 & 70 & 69 & 70 \\ \hline Friday & 68 & 70 & 69 & 71 & 72 & 69 & 70 & 70 \\ \hline \end{tabular}
State your null and alternative hypotheses relevant to the research question in symbols $H_{0}: \mu_{d} \neq 0$ vs $H_{1}: \mu_{d}=0$ $H_{0}: \mu_{1}-\mu_{2} \neq 0$ vs $H_{1}: \mu_{1}-\mu_{2}=0$ $H_{0}: \mu_{d}=0$ vs $H_{1}: \mu_{d}>0$ $H_{0}: \mu_{d}=0$ vs $H_{1}: \mu_{d} \neq 0$ $H_{0}: \mu_{1}-\mu_{2}=0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$

See Solution

Problem 352

For a certain knee surgery, a mean recovery time of 13 weeks is typical. With a new style of physical therapy, a researcher claims that the mean recovery time, $\mu$ , is less than 13 weeks. In a random sample of 32 knee surgery patients who practiced this new physical therapy, the mean recovery time is 12.8 weeks. Assume that the population standard deviation of recovery times is known to be 1.1 weeks.
Is there enough evidence to support the claim that the mean recovery time of patients who practice the new style of physical therapy is less than 13 weeks? Perform a hypothesis test, using the 0.10 level of significance. (a) State the null hypothesis $H_{0}$ and the alternative hypothesis $H_{1}$ . $\begin{array}{l} H_{0}: \square \\ H_{1}: \square \end{array}$ \begin{tabular}{ccc} $\mu$ & $\bar{x}$ & $\square^{\square}$ \\ $\square<\square$ & $\square \leq \square$ & $\square>\square$ \\ $\square \geq \square$ & $\square=\square$ & $\square \neq \square$ \\ $\times$ & $S$ \\ \hline \end{tabular} (b) Perform a $Z$ -test and find the $p$ -value.
Here is some information to help you with your Z-test. - The value of the test statistic is given by $\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$ . - The $p$ -value is the area under the curve to the left of the value of the test statistic. \begin{tabular}{l} \hline Standard Normal Distribution \\ Step 1: Select one-tailed or two-tailed. \\ One-tailed \end{tabular} Continue Submit A

See Solution

Problem 353

A sample of size $n=90$ is drawn from a normal population whose standard deviation is $\sigma=7.6$ . The sample mean is $\bar{x}=42.74$ .
Part 1 of 2 (a) Construct a $90 \%$ confidence interval for $\mu$ . Round the answer to at least two decimal places.
A $90 \%$ confidence interval for the mean is $41.42<\mu<44.06$ .
Part: $1 / 2$
Part 2 of 2 (b) If the population were not approximately normal, would the confidence interval constructed in part (a) be valid? Explain.
The confidence interval constructed in part (a) (Choose one) $\boldsymbol{\nabla}$ be valid since the sample size (Choose one) $\boldsymbol{\nabla}$ large. Skip Part Check Save For L - 2024 McGraw Hill LLC. All Rights Reserved. Terms c

See Solution

Problem 354

A machine is set to fill the small-size packages of M\&M candies with 75 candies per bag. A sample revealed three bag of 73 , four bags of 79 , three bag of 80 , and three bags of 80 . To test the hypothesis that the mean candies per bag is 77 , how many degrees of freedom are there?
Select one: a. 13 b. 4 c. 3 d. 12 e. 11

See Solution

Problem 355

Parameter estimates: \begin{tabular}{|l|c|c|c|c|c|c|} \hline Parameter & Estimate & Std. Err. & Alternative & DF & T-Stat & P-Value \\ \hline Intercept & 17248.727 & 182.09258 & $\neq 0$ & 98 & & $<0.0001$ \\ \hline Slope & -66.86088 & 4.974639 & $\neq 0$ & 98 & & $<0.0001$ \\ \hline \end{tabular}
Predicted values: \begin{tabular}{|c|c|c|} \hline X value & 95\% C.I. for mean & 95\% P.I. for new \\ \hline 41.672 & $(14376.9375,14548.063)$ & $(13808.969,15116.032)$ \\ \hline \end{tabular}
For any numeric responses, round your answers to four decimal places. a. Identify the two variables which were recorded, determining which is the predictor and which is the response. - The predictor variable is: Select an answer - The response variable is: Select an answer b. Which of the following correlation coefficients is most likely for this data? $r=$ Select an answer c. Using the correlation selected in part b, calculate $r^{2}$ and interpret this value: $\square$ $\%$ of the variability in Select an answer (6) is explained by the linear relationship with Select an answer $\square$ \% of the variability is unexplained. d. The estimated regression equation is: $y=$ $\square$ $x+$ $\square$ e. Interpret the slope in context: We expect the price of a used Toyota Camry to $\square$ when the odometer increases by $\square$ Select an answer by miles. f. Based on the scatterplot, would it be appropriate to interpret the $y$ -intercept? The $y$ -intercept can be interpreted because an odometer reading of 0 miles is plausible. The $y$ -intercept should be interpreted because a Camry with an odometer reading of 0 miles was observed in the dataset. The $y$ -intercept should not be interpreted because a Camry with an odometer of 0 miles was not observed in the dataset.

See Solution

Problem 356

Car dealers across North America use the Kelley Blue Book to help determine the value of used cars that their customers trade in when purchasing new cars. However, the Blue Book does not include the car's odometer reading as an indication of the car's value, despite the fact that this is a critical factor for usedcar buyers. It seems reasonable to believe that the more miles a used car has been driven, the less one would be willing to pay for the car. To examine this issue, a used-car dealer randomly samples 100 Toyota Camrys that were sold at auction during the past month. Each car was in top condition, three years old, and equipped with only standard features. The dealer recorded the following two variables: the selling price of the car (in dollars) and the odometer reading (in thousands of miles). StatCrunch was used to create the output below: Parameter estimates: \begin{tabular}{|l|c|c|c|c|c|c|} \hline Parameter & Estimate & Std. Err. & Alternative & DF & T-Stat & P-Value \\ \hline Intercept & 17248.727 & 182.09258 & $\neq 0$ & 98 & & $<0.0001$ \\ \hline Slope & -66.86088 & 4.974639 & $\neq 0$ & 98 & & $<0.0001$ \\ \hline \end{tabular}
Predicted values: \begin{tabular}{|c|c|c|} \hline X value & 95\% C.I. for mean & 95\% P.I. for new \\ \hline 41.672 & $(14376.9375,14548.063)$ & $(13808.969,15116.032)$ \\ \hline \end{tabular}
For any numeric responses, round your answers to four decimal places. a. Use the regression equation to provide a point estimate for the selling price of a used Toyota Camry with an odometer reading of 41,672 miles. Note: Be careful of the value you plug into the equation for the mileage - the mileage used to create the model was measured in thousands of miles. $\mathrm{y}=$ $\square$ b. The car dealership would like to estimate the average selling price for used Camrys with this odometer reading. Which interval would be the most appropriate to provide? confidence interval prediction interval c. Interpret the interval selected in part b: With $95 \%$ confidence, we estimate the
Select an answer with $\square$ miles is between $\square$ and $\square$

See Solution

Problem 357

A traffic study wants to determine if there is a relationship between behavior at a 4 -way stop sign and which direction the car turns. The study examined 425 vehicles as they approached a 4 -way stop sign keeping track of which way the car turned (left, straight, or right) and whether the car came to a complete stop or a rolling stop. Which of the following techniques would be most appropriate for analyzing the data collected? ANOVA two-sample mean, dependent samples one-sample mean one-sample proportion chi-square test
Note: You may only attempt this question 2 times and a penalty will be assessed on your second attempt. Submit Question

See Solution

Problem 358

The average cost of tuition and room and board for a small private liberal arts college is reported to be $\$ 8,500$ per term, but a financial administrator believes that the average cost is lower. A study conducted using 350 small liberal arts colleges showed that the average cost per term is $\$ 8,745$ . The population standard deviation is $\$ 1,200$ . Let $\alpha=0.05$ . What are the null and alternative hypotheses for this study?
Select one: a. $\mathrm{HO}: \mu=\$ 8,500 ; \mathrm{H1}: \mu \neq \$ 8,500$ b. $H O: \mu \geq \$ 8,500 ; H 1: \mu<\$ 8,500$ c. $H O: \mu \geq \$ 9,000 ; H 1: \mu<\$ 9,000$ d. $\mathrm{HO}: \mu \leq \$ 8,500 ; \mathrm{H} 1: \mu>\$ 8,500$ e. $H 0: \mu \leq \$ 9,000 ; H 1: \mu>\$ 9,000$
Clear my choice

See Solution

Problem 359

othesis Tests for a Question 3, 10.2.17-T HW Score: $35.88 \%, 2.51$ of 7 points Part 1 of 5 Points: 0 of 1 Save
Researchers selected 825 patients at random among those who take a certain widely-used prescription drug daily. In a clinical trial, 23 out of the 825 patients complained of flulike symptoms. Suppose that it is known that $2.5 \%$ of patients taking competing drugs complain of flulike symptoms. Is there sufficient evidence to conclude that more than $2.5 \%$ of this drug's users expenience flulike symptoms as a side effect at the $\alpha=0.01$ level of significance?
Because $\mathrm{np}_{0}\left(1-p_{0}\right)=$ $\square$ $\square$ 10 , the sample size is $\square$ $5 \%$ of the population size, and the patients in the sample $\square$ selected at random, all of the requirements for testing the hypothesis $\square$ satisfied. (Round to one decimal place as needed.)

See Solution

Problem 360

A drug, which is used for treating cancer, has potentially dangerous side effects if it is taken in doses which are larger than the required dosage for the treatment. The tablet should contain 54.94 mg and the standard deviation should be 0.04 . 25 tablets are randomly selected and the amount of drug in each tablet is measured. The sample has a mean of 54.945 mg and a variance of 0.0004 mg . Does the data suggest at $\alpha=0.025$ that the tablets vary by less than the desired amount?
Step 2 of 5 : Determine the critical value(s) of the test statistic. If the test is two-tailed, separate the values with a comma. Round your answer to three decimal places.
Answer How to enter your answer (opens in new window) Tables Keypad Keyboard Shortcuts Previouls step aniswers Submit Answer - 2024 Hawkes Learning

See Solution

Problem 361

Submit Answer
5. [-/12.5 Points] DETAILS MY NOTES BBUNDERSTAT13 8.1.021. How much a customer buys is a direct result of how much time they spend in a store. A study of average shopping times in a large chain store gave the following information: Shopping alone: 9 min. Shopping with a family: 31 min. Suppose you want to set up a statistical test to challenge the claim that people who shop alone spend, on average, 19 minutes shopping in a store. (a) What would you use for the null and alternate hypotheses (in min) if you believe the average shopping time is less than 19 minutes? (Enter != for as needed.) Ho H1 Is this a right-tailed, left-tailed, or two-tailed test? O right-tailed two-tailed left-tailed (b) What would you use for the null and alternate hypotheses (in min) if you believe the average shopping time is different from 19 minutes? (Enter != for as needed.) Ho₁ H₁i Is this a right-tailed, left-tailed, or two-tailed test? O right-tailed O two-tailed O left-tailed

See Solution

Problem 362

Hypothesis Tests for a Question 6, 10.3.15-T HW Score: $54.58 \%, 4.91$ of 9 points Part 4 of 4 Points: 0 of 1 Save
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 $9.02 \mathrm{~cm}^{3}$ . An analysis of the sample data revealed that the hippocampal volume is approximately normal with no outliers and $\overline{\mathrm{x}}=8.03 \mathrm{~cm}^{3}$ and $\mathrm{s}=0.8 \mathrm{~cm}^{3}$ . Conduct the appropriate test at the $\alpha=0.01$ level of significance. $\begin{array}{l} H_{0}: \mu=y .02 \\ H_{1}: \mu<9.02 \end{array}$ (Type integers or decimals. Do not round.) Identify the $t$ -statistic. $t_{0}=-4.11$ (Round to two decimal places as needed) Identify the P -value. $P$ -value $=001$ (Round to three decimal places as needed.) Make a conclusion regarding the hypothesis. Fail to reject the null hypothesis. There $\square$ sufficient evidence to claim that the mean hippocampal volume is $\square$ equal to $\square$ $\mathrm{cm}^{3}$

See Solution

Problem 363

Question 25 of 33 (1 point)
Sleep apnea: Sleep apnea is a disorder in which there are pauses in breathing during sleep. People with this condition must wake up frequently to breathe. In a sample of 426 people aged 65 and over, 120 of them had sleep apnea.
Part 1 of 3 (a) Find a point estimate for the population proportion of those aged 65 and over who have sleep apnea. Round the answer to three decimal places.
The point estimate for the population proportion of those aged 65 and over who have sleep apnea is 0.282 .
Part 2 of 3 (b) Construct a $80 \%$ confidence interval for the proportion of those aged 65 and over who have sleep apnea. Round the answer to three decimal places.
A $80 \%$ confidence interval for the proportion of those aged 65 and over who have sleep apnea is $0.254<p<0.310$ .
Part: $2 / 3$
Part 3 of 3 (c) In another study, medical researchers concluded that more than $16 \%$ of elderly people have sleep apnea. Based on the confidence interval, does it appear that more than $16 \%$ of people aged 65 and over have sleep apnea? Explain. Since all of the values in the confidence interval are (Choose one) $\boldsymbol{\nabla} 0.16$ , the confidence interval (Choose one) $\nabla$ contradict the claim.

See Solution

Problem 364

A research scientist wants to know how many times per hour a certain strand of bacteria reproduces. The mean is found to be 11.4 reproductions and the population standard deviation is known to be 2.4. If a sample of 262 was used for the study, construct the $80 \%$ confidence interval for the true mean number of reproductions per hour for the bacteria. Round your answers to one decimal place.
Answer How to enter your answer (opens in new window) Tables Keypad Keyboard Shortcuts
Lower endpoint: $\square$ Upper endpoint: $\square$

See Solution

Problem 365

A $95 \%$ confidence interval for $\mu$ is computed to be $(1.60,4.15)$ . For each of the following hypotheses, state whether $H_{0}$ will be rejected at 0.05 level. Part: $0 / 4$
Part 1 of 4 $H_{0}: \mu=4$ versus $H_{1}: \mu \neq 4$ Since the $95 \%$ confidence interval (Choose one) $\boldsymbol{\nabla} \mu_{0}$ , then $H_{0}$ (Choose one) $\boldsymbol{\nabla}$ be rejected at the 0.05 level.

See Solution

Problem 366

If $P=0.04$ , which of the following is the best conclusion? If $H_{0}$ is false, the probability of obtaining a test statistic as extreme as or more extreme than the one actually observed is 0.04 . The probability that $H_{0}$ is true is 0.04 . The probability that $H_{0}$ is false is 0.04 . If $H_{0}$ is true, the probability of obtaining a test statistic as extreme as or more extreme than the one actually observed is 0.04 .

See Solution

Problem 367

Question 10 of 15, Step 2 of 3 Correct
The board of a major credit card company requires that the mean wait time for customers when they call customer service is at most 3.50 minutes. To make sure that the mean wait time is not exceeding the requirement, an assistant manager tracks the wait times of 41 randomly selected calls. The mean wait time was calculated to be 3.94 minutes. Assuming the population standard deviation is 1.30 minutes, is there sufficient evidence to say that the mean wait time for customers is longer than 3.50 minutes with a $95 \%$ level of confidence? Step 2 of 3 : Compute the value of the test: statistic. Round your answer ter two decimal places.
Answer Tables Keypad Keyboard Shortcuts

See Solution

Problem 368

The frequency distribution shows the results of 200 test scores. Are the test scores normally distributed? Use $\alpha=0.05$ . Complete parts (a) through (e) \begin{tabular}{|l|c|c|c|c|c|} \hline Class boundaries & $49.5-58.5$ & $58.5-67.5$ & $67.5-76.5$ & $76.5-85.5$ & $85.5-94.5$ \\ \hline Frequency, $f$ & 20 & 61 & 79 & 35 \\ \hline \end{tabular}
Using a chi-square goodness-of-fit test, you can decide, with some degree of certainty, whether a variable is normally distributed. In all chi-square tests for normality, the null and alternative hypotheses are as follows. $H_{0}$ . The test scores have a normal distribution. $H_{\mathrm{a}}$ . The test scores do not have a normal distribution. To determine the expected frequencles when performing a chi-square test for normality, first find the mean and standard deviation of the frequency distribution. Then, use the mean and standard deviation to compute the $z$ -score for each class boundary. Then, use the z-scores to calculate the area under the standard normal curve for each class. Multiplying the resulting class areas by the sample size yields the expected frequency for each class. (a) Find the expected frequencies \begin{tabular}{|l|c|c|c|c|c|} \hline Class boundaries & $49.5-58.5$ & $58.5-67.5$ & $67.5-76.5$ & $76.5-85.5$ \\ \hline Frequency, $f$ & 20 & 61 & 79 & 35 & 35 \\ \hline Expected frequency & $\square$ & $\square$ & $\square$ & $\square$ \\ \hline \end{tabular} (Round to the nearest integer as needed)

See Solution

Problem 369

For the following, indicate whether a confidence interval for a proportion or mean should be constructed to estimate the variable of interest. Justify your response. Does chewing your food for a longer period of time reduce one's caloric intake of food at dinner? A researcher requires a sample of 75 healthy males to chew their food twice as long as they normal do. The researcher then records the calorie consumption at dinner.
The confidence interval for a mean $\square$ should be constructed because the variable of interest is an individual's reduction in caloric intake, $\square$ which is a $\square$ variable. quantitative qualitative

See Solution

Problem 370

The Food and Nutrition Board of the National Academy of Sciences reports that the mean daily sodium intake should not exceed $3,100 \mathrm{mg}$ . In a study of sodium intake, a sample of $\mathrm{U} . \mathrm{S}$ residents was found to have a mean daily sodium intake of 3,700 and a standard deviation of $1,050 \mathrm{mg}$ . based on a sample of size 120. The reserachers were interested in determining whether the mean daily sodium intake for U.S residents exceeded the maximum recommended level.
What is the value of the test statistic? 0.571 $-6.26$ $-0.571$ 6.26

See Solution

Problem 371

Weights of Vacuum Cleaners Upright vacuum cleaners have either a hard body type or a soft body type. Shown are the weights in pounds of a random sample of each type. Find the $90 \%$ confidence interval for the difference of the means. Assume the variables are normally distributed and the variances are unequal. Round the sample mean and sample standard deviation to two decimal places and the final answers to at least one decimal place. \begin{tabular}{llll|llll} \multicolumn{4}{c|}{ Hard body types } & \multicolumn{4}{c}{ Soft body types } \\ \hline 21 & 17 & 17 & 20 & 24 & 13 & 11 & 13 \\ 16 & 17 & 15 & 20 & 12 & 15 & & \\ 23 & 16 & 17 & 17 & & & & \\ 13 & & & & & & \end{tabular}
Send data to Excel
Use $\mu_{1}$ for the mean weight of the hard body type. $\square<\mu_{1}-\mu_{2}<\square$

See Solution

Problem 372

A tire manufacturer tested 100 randomly selected tires to evaluate the risk involved in its 50,000-kilometer warranty. A tire passed the test if it did not show a defect during a 50,000kilometer simulation. \begin{tabular}{|l|c|} \hline Test result & Tires \\ \hline pass & 85 \\ \hline fail & 15 \\ \hline \end{tabular}
Find a $95 \%$ confidence interval for $p$ , the proportion of tires that would pass the test. Round your answers to the nearest thousandth. $\square$ $<p<$ $\square$

See Solution

Problem 373

To encourage water conservation, a newspaper is publishing an article about xeriscaping, a method of landscaping that requires less water. For the article, a newspaper reporter contacted 325 randomly chosen homeowners, and asked them what kind of lawns they have. \begin{tabular}{|l|c|} \hline Lawn type & Homeowners \\ \hline natural grass & 43 \\ \hline artificial turf & 275 \\ \hline rock & 2 \\ \hline no lawn area & 5 \\ \hline \end{tabular}
Find a $95 \%$ confidence interval for $p$ , the proportion of homeowners with natural grass lawns. Round your answers to the nearest thousandth. $\square$ $<p<$ $\square$

See Solution

Problem 374

Meth Narifroticin seced from a binomia popultason with unknown probability of success $p$ . The computed value of $\hat{p}$ is 0.64 ertaceste $=816$ tefretconolionis velater: $=-209$ coerte orical: reserecitar 2 sm rab har canclion of macratem
Anocomechaswers

See Solution

Problem 375

10.
The accompanying table shows the enrollment of a preschool from 1980 through 2000. Write a linear regression equation to model the data in the table. \begin{tabular}{|c|c|} \hline Year $(x)$ & Enrollment $(x)$ \\ \hline 1980 & 14 \\ \hline 1985 & 20 \\ \hline 1990 & 22 \\ \hline 1995 & 28 \\ \hline 2000 & 37 \\ \hline \end{tabular}
What is the correlation coefficient?
Interpret the sign of the correlation coefficient about the line of best fit.

See Solution

Problem 376

Gentle Ben is a Morgan horse at a Colorado dude ranch. Over the past 8 weeks, a veterinarian took the following glucose readings from this horse (in mg/100 ml ). \begin{tabular}{|l|l|l|l|l|l|l|l|} \hline 91 & 86 & 84 & 105 & 99 & 112 & 82 & 89 \\ \hline \end{tabular}
The sample mean is $\bar{x}=93.50$ . Let $x$ be a random variable representing glucose readings taken from Gentle Ben. We may assume that $x$ has a normal distribution, and we know from past experience that $\sigma=12.5$ . The mean glucose level for horses should be $\mu=85 \mathrm{mg} / 100 \mathrm{ml}$ . + Do these data indicate that Gentle Ben has an overall average glucose level higher than 85 ? Use $\alpha=0.05$ .

See Solution

Problem 377

Test if the mean score of sober subjects is 35 using scores: mean = 41, std dev = 3.7, n = 20, significance = 0.01.
Test if the proportion of women (65%) favoring gun control is higher than men (60%) using samples of 360 women and 220 men, significance = 0.05.

See Solution

Problem 378

Commuters and parking spaces data: a) Find the correlation coefficient. b) Find critical values for $r$ . c) Check for significant correlation at 0.05 level. d) State the conclusion.
CPI and subway fare data: a) Find regression equation with CPI as $x$ . b) Predict subway fare for CPI = 182.5.

See Solution

Problem 379

Construct a 95% confidence interval for the standard deviation of 23 PG-rated movies with mean 120.8 and SD 22.9. Show steps.

See Solution

Problem 380

A researcher tests if a diet lowers blood pressure ( $\mathrm{mm} \mathrm{Hg}$ ) for those with high blood pressure. Use $\alpha = 0.01$ to compare means from treatment (n=35, $\bar{x}=189.1$ , s=38.7) and control (n=28, $\bar{x}=203.7$ , s=39.2).
a) State null and alternative hypotheses. b) Calculate test statistic and $\mathrm{p}$ -value. c) Make a decision. d) Interpret conclusion. e) Find and interpret the 99% confidence interval.

See Solution

Problem 381

1. Given commuter and parking space data, find the correlation coefficient, critical values for $r$ , and check significance at 0.05.
2. For CPI and subway fare data, find the regression equation with CPI as $x$ and predict fare for CPI = 182.5.

See Solution

Problem 382

Test if the mean score of sober subjects is 35 using a sample mean of 41, SD of 3.7, at 0.01 significance level.
Also, test if the proportion of women (65%) favoring stricter gun control is higher than men (60%) at 0.05 significance level.

See Solution

Problem 383

Construct a confidence interval for LDL cholesterol changes: mean = 3.2, SD = 18.6. What is the best point estimate?

See Solution

Problem 384

A chemist studies how carbon atoms in hydrocarbons affect energy release during combustion. Estimate $\theta_{0}$ and $\theta_{1}$ for $y = \theta_{0} + \theta_{1} x$ . Options: $\theta_{0}=-569.6, \theta_{1}=530.9$ ; $\theta_{0}=-1780.0, \theta_{1}=530.9$ ; $\theta_{0}=-569.6, \theta_{1}=-530.9$ ; $\theta_{0}=-1780.0, \theta_{1}=-530.9$ .

See Solution

Problem 385

Find Sharpe's measure for Ella's stock with beta 1.34, std. dev. 16.4%, return 14.8%, market risk premium 8.5%, market return 12.0%.

See Solution

Problem 386

How many people were unemployed in Africa in 2014 if the number is expected to reach 61 million in 2024, a $24.5 \%$ increase? Round to the nearest million.

See Solution

Problem 387

Construct a confidence interval for garlic's effect on LDL cholesterol. Mean change: 3.2, SD: 18.6. Find point estimate and $95\%$ CI.

See Solution

Problem 388

1. Given commuter and parking space data, find the correlation coefficient, critical values for $r$ , and test for significance at 0.05.
2. For CPI and subway fare, find the regression equation with CPI as $x$ and predict fare for CPI = 182.5.

See Solution

Problem 389

Explain $k$ -fold cross-validation implementation and its pros/cons vs. validation set approach and LOOCV.

See Solution

Problem 390

Suppose the results indicate that the null hypothesis should be rejected; thus, it is possible that a type I error has been committed.
Given the type of error made in this situation, what could researchers do to reduce the risk of this error? Instead of a .05 significance level, choose a .01 significance level Increase the sample size.

See Solution

Problem 391

ervals Quiz Question 7 of 13 This quiz: 13 point(s) possible This question: 1 point(s) possible Submit quiz
Here are summary statistics for randomly selected weights of newborn girls: $\mathrm{n}=36, \overline{\mathrm{x}}=3216.7 \mathrm{~g}, \mathrm{~s}=688.5 \mathrm{~g}$ . Use a confidence level of $95 \%$ to complete parts (a) through (d) below. a. Identify the critical value $\mathrm{t}_{\alpha / 2}$ used for finding the margin of error. $\mathrm{t}_{\alpha / 2}=\square$ (Round to two decimal places as needed.) b. Find the margin of error. $E=\square g$ (Round to one decimal place as needed.) c. Find the confidence interval estimate of $\mu$ . $\square$ $g<\mu<$ $\square$ $g$ (Round to one decimal place as needed.) d. Write a brief statement that interprets the confidence interval. Choose the correct answer below.

See Solution

Problem 392

If $n=12, \bar{x}=50$ , and $s=11$ , construct a confidence interval at a $98 \%$ confidence level. Assume the data came from a normally distributed population. $\square$ $\square$ Give your answers to one decimal place. Can we conclude that the population mean is at least 42.67 ? No Yes
Can we conclude that the population mean is no more than 60.33 ? Yes No

See Solution

Problem 393

(Note: For any subtraction, do A - B.) 200 out of a sample 333 students in State A own an iPhone while 28 out 44 in State B had an iPhone. We'd like to know are the proportions of students with iPhones different in the two states.
Test Statistic: $\square$ According to your tables, the P -value is approximately: $P$ -value $\approx$ $\square$ Is there sufficient evidence to conclude that a different proportion of students own iPhones in the two states? Choose one $\square$ Find the $99 \%$ confidence interval for the difference in population proportions.

See Solution

Problem 394

$\text{A very large company with 8000 employees and campuses in San Francisco (SF) and San Jose (SJ) wants to know if the proportion of employees who drive to work is different at the two campuses. Forty percent of employees work in San Francisco and the others work in San Jose. A voluntary email survey was sent out and 20\% of the SF employees responded and 50\% of SJ employees responded. The survey found that, out of those who responded, 45\% at the SF campus drove and 50\% drove at SJ.}$
$\text{Test Statistic: } \square$
$\text{According to your tables, the P-value is approximately:}$
$P-\text{value} \approx \square$
$\text{Is there sufficient evidence to conclude that a different proportion drive at the two campuses?}$
$\text{Choose one}$

See Solution

Problem 395

For each problem, complete the hypothesis testing procedure using ANOVA. Follow the general outline for hypothesis testing presented in lecture and the assigned reading and clearly label each of the following: 1) state the research hypothesis; 2) complete hypothesis testing steps 1-4 for steps specific to ANOVA); and 3) state the conclusions about the research hypothesis justified by the ANOVA, including whether any post hoc tests are required. Finally, make certain you provide an ANOVA summary table as the last piece of work you do in Step 3-the calculations -to help you become familiar with reading and understanding what the ANOVA produces.
1. A company which manufactures major electrical appliances maintains a school for training the repairpersons who are to be employed in the service departments of dealers across the country. The teachers of the course have been arguing forever about the most effective way to train people, and finally decide to conduct an experiment in which three different training methods will be compared. 45 people are randomly assigned to 3 groups of 15 , and each group is taught by a different method. At the end of the course each person is given a set of practical problems to solve, and a total proficiency score is obtained for each trainee. You work in the research department, and now it is your responsibility to determine if there are any significant differences between training methods. If you reject your null hypothesis, please conduct post hoc Tukey HSD tests to determine which means are significantly different.
Given: $\begin{array}{l} \mathrm{T}_{1}=354 ; \mathrm{T}_{2}=220 ; \mathrm{T}_{3}=256 \\ \mathrm{G}=830 \\ \sum \mathrm{X}^{2}=16,282 \\ \mathrm{k}=3 \\ \mathrm{n}_{1}=\mathrm{n}_{2}=\mathrm{n}_{3}=15 ; \mathrm{N}=45 \end{array}$ \begin{tabular}{|l|l|l|l|l|l|} \hline Source & SS & df & MS $\left(\mathrm{s}^{2}\right)$ & F & p \\ \hline Between Treatments & & & & & \\ \hline Within Treatments & 331.867 & & & & \\ \hline Total & & & & & \\ \hline \end{tabular}

See Solution

Problem 396

Evaluate $z=\frac{\hat{p}_{1}-\hat{p}_{2}-0}{S E}$ for $\hat{p}_{1}=0.413, \hat{p}_{2}=0.343$ , and $S E=0.165$ $z=$ $\square$ (Round to two decimal places as needed.)

See Solution

Problem 397

Suppose a sample size of 87 was used to conduct a right tailed hypothesis test for a single population mean. The null hypothesis was $\mathrm{H}_{0}: \mu=45$ and the sample mean was 50 and the p -value was 0.03 . Then there is a $3 \%$ chance that 87 randomly selected individuals from a population with a mean of 45 and the same standard deviation would have a sample mean greater than 50. true false

See Solution

Problem 398

5 pts 1 Details
The numbers of online applications from simple random samples of college applications for 2004 and for the 2008 were taken. In 2004, out of 658 applications, 290 of them were completed online. In 2008, out of 236 applications, 118 of them were completed online. Test the claim that the proportion of online applications in 2004 was equal to the proportion of online applications in 2008 at the .025 significance level.
Claim: Select an answer which corresponds to Select an answerv Opposite: Select an answer which corresponds to Select an answer
The test is: Select an answer The test statistic is: $z=$ $\square$ (to 2 decimals)
The critical value is: $z=$ $\square$ (to 2 decimals)
Based on this we: Select an answer $\square$ Conclusion There Select an answer appear to be enough evidence to support the claim that the proportion of online applications in 2004 was equal to the proportion of online applications in 2008.

See Solution

Problem 399

Rhino viruses typically cause common colds. In a test of the effectiveness of echinacea, 36 of the 42 subjects treated with echinacea developed rhinovirus infections. In a placebo group, 83 of the 99 subjects developed rhinovirus infections. Use a 0.01 significance level to test the claim that echinacea has an effect on rhinovirus infections. Complete parts (a) through (b) below. a. Test the claim using a hypothesis test.
Consider the first sample to be the sample of subjects treated with echinacea and the second sample to be the sample of subjects treated with a placebo. What are the null and alternative hypotheses for the hypothesis test? A. $H_{0}: p_{1}=p_{2}$ B. $H_{0}: p_{1} \leq p_{2}$ C. $H_{0}: p_{1}=p_{2}$ $H_{1}: p_{1} \neq p_{2}$ $H_{1}: p_{1} \neq p_{2}$ D. $H_{0}: p_{1} \geq p_{2}$ E. $H_{0}: p_{1} \neq p_{2}$ $H_{1}: p_{1} \neq p_{2}$ $H_{1}: p_{1}=p_{2}$ F. $H_{0}: p_{1}=p_{2}$ $H_{1}: p_{1}>p_{2}$
Identify the test statistic. $z=$ $\square$ (Round to two decimal places as needed.) Identify the P -value. P -value $=$ $\square$ (Round to three decimal places as needed.) What is the conclusion based on the hypothesis test? The $P$ -value is $\square$ the significance level of $\alpha=0.01$ , so $\square$ the null hypothesis. There $\square$ sufficient evidence to support the

See Solution

Problem 400

A quality control inspector finds that a sample of 60 manufactured parts has an average length of 12.5 cm . The population standard deviation of part length is 1.2 cm . Determine a $99 \%$ confidence interval for the population mean length of the parts.

See Solution

1 2 3 4 5 6 7 ...14

Statistical Inference

Problem 301

Problem 302

Problem 303

Problem 304

Problem 305

Problem 306

Problem 307

Problem 308

Problem 309

Problem 310

Problem 311

Problem 312

Problem 313

Problem 314

Problem 315

Problem 316

Problem 317

Problem 318

Find the critical tscore to three decimal places when the Confidence Level is 93%93 \%93% and the sample size is 32 A. 2.245 B. 2.891 C. 1.682 D. 1.877

Problem 319

Problem 320

Problem 321

Problem 322

Problem 323

ListenProvide the null and alternative hypotheses to test whether μ\muμ is less than 50. a) H0:μ≥50H_{0}: \mu \geq 50H0​:μ≥50 vs. Ha:μ<50H_{a}: \mu<50Ha​:μ<50 b) H0:μ≤50H_{0}: \mu \leq 50H0​:μ≤50 vs. Ha:μ>50H_{a}: \mu>50Ha​:μ>50

Problem 324

Problem 325

Problem 326

Problem 327

Problem 328

Problem 329

Problem 330

Problem 331

Problem 332

Problem 333

Problem 334

Problem 335

Problem 336

6. A medical researcher claims that 5%5 \%5% of children under 18 years of age have asthma. In a random sample of 250 children under 18 years of age, 9.6%9.6 \%9.6% say they have asthma. At α=0.01\alpha=0.01α=0.01, is there enough evidence to reject the researcher's claim?

Problem 337

Problem 338

Problem 339

Problem 340

Problem 341

Problem 342

Problem 343

Problem 344

Problem 345

Problem 346

Problem 347

Problem 348

Problem 349

Problem 350

Problem 351

Problem 352

Problem 353

Problem 354

Problem 355

Problem 356

Problem 357

Problem 358

Problem 359

Problem 360

Problem 361

Problem 362

Problem 363

Problem 364

Problem 365

Problem 366

Problem 367

Problem 368

Problem 369

Problem 370

Problem 371

Problem 372

Problem 373

Problem 374

Meth Narifroticin seced from a binomia popultason with unknown probability of success ppp. The computed value of p^\hat{p}p^​ is 0.64 ertaceste =816=816=816 tefretconolionis velater: =−209=-209=−209 coerte orical: reserecitar 2 sm rab har canclion of macratemAnocomechaswers

Problem 375

Find the critical tscore to three decimal places when the Confidence Level is $93 \%$ and the sample size is 32 A. 2.245 B. 2.891 C. 1.682 D. 1.877

Listen
Provide the null and alternative hypotheses to test whether $\mu$ is less than 50. a) $H_{0}: \mu \geq 50$ vs. $H_{a}: \mu<50$ b) $H_{0}: \mu \leq 50$ vs. $H_{a}: \mu>50$

6. A medical researcher claims that $5 \%$ of children under 18 years of age have asthma. In a random sample of 250 children under 18 years of age, $9.6 \%$ say they have asthma. At $\alpha=0.01$ , is there enough evidence to reject the researcher's claim?

Meth Narifroticin seced from a binomia popultason with unknown probability of success $p$ . The computed value of $\hat{p}$ is 0.64 ertaceste $=816$ tefretconolionis velater: $=-209$ coerte orical: reserecitar 2 sm rab har canclion of macratem
Anocomechaswers