Math  /  Data & Statistics

QuestionThe weights of four randomly and independently selected bags of tomatoes labeled 5 pounds were found to be 5.2,4.9,5.25.2,4.9,5.2, and 5. 맘 Assume Normality. a. Using a two-sided alternative hypothesis, should you be able to reject the hypothesis that the population mean is 5pounds using a significance level of 0.05 Why or why not? The confidence interval is reported here: I am 95\%onfident the population mean is between 4.8and 5.3pounds. b. Now test the hypothesis that the population mean is not 5 pounds. Use a significance level of 0.05 . b. Determine the null and alternative hypotheses. Choose the correct answer below. A. H0:μ=5\mathrm{H}_{0}: \mu=5 B. H0:μ=5H_{0}: \mu=5 C. H0:μ>5\mathrm{H}_{0}: \mu>5 Ha:μ<5H_{a}: \mu<5 Ha:μ>5H_{a}: \mu>5 Ha:μ5H_{a}: \mu \leq 5 D. H0:μ5Ha:μ=5\begin{array}{l} H_{0}: \mu \neq 5 \\ H_{a}: \mu=5 \end{array} E. H0:μ=5Ha:μ5\begin{array}{l} H_{0}: \mu=5 \\ H_{a}: \mu \neq 5 \end{array} F. H0:μ<5\mathrm{H}_{0}: \mu<5 Ha:μ5H_{a}: \mu \geq 5
Find the test statistic. t=\mathrm{t}= \square (Round to two decimal places as needed.)

Studdy Solution

STEP 1

1. The weights of the bags are normally distributed.
2. We are using a significance level of α=0.05 \alpha = 0.05 .
3. The sample consists of four weights: 5.2,4.9,5.2,5.0 5.2, 4.9, 5.2, 5.0 .

STEP 2

1. Analyze the given confidence interval for part (a).
2. Determine the null and alternative hypotheses for part (b).
3. Calculate the test statistic for part (b).

STEP 3

Analyze the given confidence interval:
The confidence interval is (4.8,5.3) (4.8, 5.3) .
Since the hypothesized population mean of 5 pounds is within the 95% confidence interval, we do not have enough evidence to reject the null hypothesis at the 0.05 significance level.

STEP 4

Determine the null and alternative hypotheses:
The correct choice for a two-sided test is:
E. H0:μ=5Ha:μ5\begin{array}{l} H_{0}: \mu=5 \\ H_{a}: \mu \neq 5 \end{array}

STEP 5

Calculate the test statistic:
1. Calculate the sample mean (xˉ\bar{x}):
xˉ=5.2+4.9+5.2+5.04=5.075\bar{x} = \frac{5.2 + 4.9 + 5.2 + 5.0}{4} = 5.075
2. Calculate the sample standard deviation (ss):
s=(5.25.075)2+(4.95.075)2+(5.25.075)2+(5.05.075)241s = \sqrt{\frac{(5.2 - 5.075)^2 + (4.9 - 5.075)^2 + (5.2 - 5.075)^2 + (5.0 - 5.075)^2}{4 - 1}}
s=0.015625+0.030625+0.015625+0.0056253=0.06753=0.0225=0.15s = \sqrt{\frac{0.015625 + 0.030625 + 0.015625 + 0.005625}{3}} = \sqrt{\frac{0.0675}{3}} = \sqrt{0.0225} = 0.15
3. Calculate the test statistic (tt):
t=xˉμs/n=5.07550.15/4=0.0750.075=1.00t = \frac{\bar{x} - \mu}{s / \sqrt{n}} = \frac{5.075 - 5}{0.15 / \sqrt{4}} = \frac{0.075}{0.075} = 1.00
The test statistic is:
t=1.00t = \boxed{1.00}

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