For x‾1=40,σ1=8.3,n1=40,x‾2=39,σ2=5.8\overline{x}_1=40,\sigma_1=8.3,n_1=40,\overline{x}_2=39,\sigma_2=5.8, & n2=50n_2=50, perform a hypothesis test to test the claim that μ1>μ2\mu_1>\mu_2 for α=0.05\alpha=0.05.

P − v a l u e > α P-value>\alpha P − v a l u e > α ; fail to reject H 0 H_0 H 0 since there is not enough evidence to suggest μ 1 > μ 2 \mu_1>\mu_2 μ 1 > μ 2 .

For x ‾ 1 = 40 , σ 1 = 8.3 , n 1 = 40 , x ‾ 2 = 39 , σ 2 = 5.8 \overline{x}_1=40,\sigma_1=8.3,n_1=40,\overline{x}_2=39,\sigma_2=5.8 , & n 2 = 50 n_2=50 , perform a hypothesis test to test the claim that μ 1 > μ 2 \mu_1>\mu_2 for α = 0.05 \alpha=0.05 .

Welcome back. For this problem, we've been given two population standard deviations, and for a sample from each of those populations, we have a mean and a sample size. We'd like to perform a hypothesis test to test the claim that mu one is greater than mu two for alpha equals point zero five. So let's start by writing our hypothesis. The null hypothesis is that both means are the same, mu one is equal to mu two. And the alternative hypothesis is what we're trying to test, which in this case is that mu one is greater than mu two. Now that we have our hypothesis, let's turn to our t I 80 fours and run the test. We'll click on the stat button, scroll two to the right to the test menu, and click on three. That's step one. All set. We wanna enter in statistics and parameters, so we'll scroll over to stats and click enter. Then we'll put in the information we have from the problem. Sigma one is 8.3, and sigma two is 5.8. Then we'll enter in the information for our first sample, which has a mean of 40 and a size of 40, and the information for the second sample, which has a mean of 39 and a size of 50. Once that's all set, we'll select the correct alternative hypothesis, which is that mu one is greater than mu two. So we want greater than mu two selected. Then we'll scroll down to calculate and click enter. That's step two, all set. And we'll end up getting the p value in our viewing window, which is approximately 0.26. We wanna compare that to 0.05, which is our alpha value. We can see our p value is greater than alpha, so we fail to reject our null hypothesis, which means that there is not enough evidence to suggest that the alternative hypothesis is true, or that mu one is greater than mu two. Great work.

For x ‾ 1 = 40 , σ 1 = 8.3 , n 1 = 40 , x ‾ 2 = 39 , σ 2 = 5.8 \overline{x}_1=40,\sigma_1=8.3,n_1=40,\overline{x}_2=39,\sigma_2=5.8 , & n 2 = 50 n_2=50 , perform a hypothesis test to test the claim that μ 1 > μ 2 \mu_1>\mu_2 for α = 0.05 \alpha=0.05 .

P − v a l u e > α P-value>\alpha P − v a l u e > α ; fail to reject H 0 H_0 H 0 since there is not enough evidence to suggest μ 1 > μ 2 \mu_1>\mu_2 μ 1 > μ 2 .

P − v a l u e < α P-value<\alpha P − v a l u e < α ; fail to reject H 0 H_0 H 0 since there is not enough evidence to suggest μ 1 > μ 2 \mu_1>\mu_2 μ 1 > μ 2 .

P − v a l u e > α P-value>\alpha P − v a l u e > α ; reject H 0 H_0 H 0 since there is enough evidence to suggest μ 1 > μ 2 \mu_1>\mu_2 μ 1 > μ 2 .

P − v a l u e < α P-value<\alpha P − v a l u e < α ; reject H 0 H_0 H 0 since there is enough evidence to suggest μ 1 > μ 2 \mu_1>\mu_2 μ 1 > μ 2 .

10. Hypothesis Testing for Two Samples

Two Means - Known Variance

Statistics

10. Hypothesis Testing for Two Samples

Two Means - Known Variance

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Multiple Choice

For ${\overline{x}}_{1} = 40, σ_{1} = 8.3, n_{1} = 40, {\overline{x}}_{2} = 39, σ_{2} = 5.8$ , & $n sub 2 equals 50$ , perform a hypothesis test to test the claim that $mu sub 1 is greater than mu sub 2$ for $alpha equals 0.05$ .

$P-value<\alpha$ ; fail to reject $H_0$ since there is not enough evidence to suggest $\mu_1>\mu_2$ .

$P-value>\alpha$ ; reject $H_0$ since there is enough evidence to suggest $\mu_1>\mu_2$ .

$P-value<\alpha$ ; reject $H_0$ since there is enough evidence to suggest $\mu_1>\mu_2$ .

$P-value>\alpha$ ; fail to reject $H_0$ since there is not enough evidence to suggest $\mu_1>\mu_2$ .

Verified step by step guidance

Identify the null and alternative hypotheses. Here, the claim is that $ \mu_1 > \mu_2 $, so set $ H_0: \mu_1 \leq \mu_2 $ and $ H_a: \mu_1 > \mu_2 $.

Calculate the test statistic for the difference between two means with known variances using the formula: \[ z = \frac{(\overline{x}_1 - \overline{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \] where $ \mu_1 - \mu_2 = 0 $ under the null hypothesis.

Substitute the given values $ \overline{x}_1 = 40 $, $ \overline{x}_2 = 39 $, $ \sigma_1 = 8.3 $, $ \sigma_2 = 5.8 $, $ n_1 = 40 $, and $ n_2 = 50 $ into the formula to compute the z-score.

Find the p-value corresponding to the calculated z-score for a right-tailed test (since $ H_a $ is $ \mu_1 > \mu_2 $) using the standard normal distribution.

Compare the p-value to the significance level $ \alpha = 0.05 $. If $ \text{p-value} < \alpha $, reject $ H_0 $; otherwise, fail to reject $ H_0 $.

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Textbook Question

In Exercises 11–14, test the claim about the difference between two population means and at the level of significance . Assume the samples are random and independent, and the populations are normally distributed.

Claim: μ1>μ2; α=0.10

Population statistics:σ1=40 and σ2=15

Sample Statistics: x̅1=500, n1=100, x̅2=495, n2=75

Textbook Question

Testing the Difference Between Two Means In Exercises 15–24, (a) identify the claim and state Ho and Ha, (b) find the critical value(s) and identify the rejection region(s), (c) find the standardized test statistic z, (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. Assume the samples are random and independent, and the populations are normally distributed.

Bed-in-a-Box To compare customer satisfaction with mattresses that are delivered compressed in a box and traditional mattresses, a researcher randomly selects 30 ratings of mattresses in boxes and 30 ratings of traditional mattresses. The mean rating of mattresses in boxes is 68.7 out of 100. Assume the population standard deviation is 6.6. The mean rating of traditional mattresses is 70.9 out of 100. Assume the population standard deviation is 5.6. At α=0.01, can the researcher support the claim that the mean rating of traditional mattresses is greater than the mean rating of mattresses in a box? (Adapted from Consumer Reports)

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Textbook Question

Repair Costs: Washing Machines You want to buy a washing machine, and a salesperson tells you that the mean repair costs for Model A and Model B are equal. You research the repair costs. The mean repair cost of 24 Model A washing machines is \$208. Assume the population standard deviation is \$18. The mean repair cost of 26 Model B washing machines is \$221. Assume the population standard deviation is \$22. At α=0.01, can you reject the salesperson’s claim?

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Textbook Question

Braking Distances To compare the dry braking distances from 60 to 0 miles per hour for two makes of automobiles, a safety engineer conducts braking tests for 16 compact SUVs and 11 midsize SUVs. The mean braking distance for the compact SUVs is 131.8 feet. Assume the population standard deviation is 5.5 feet. The mean braking distance for the midsize SUVs is 132.8 feet. Assume the population standard deviation is 6.7 feet. At α=0.10 , can the engineer support the claim that the mean braking distances are different for the two categories of SUVs? (Adapted from Consumer Reports)

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Textbook Question

Claim: μ1=μ2; α=0.1

Population statistics:σ1=3.4 and σ2=1.5

Sample Statistics: x̅1=16, n1=29, x̅2=14, n2=28

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Textbook Question

Claim: μ1>μ2; α=0.10

Population statistics:σ1=40 and σ2=15

Sample Statistics: x̅1=500, n1=100, x̅2=495, n2=75

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