For x‾1=41,Sx1=8.3,n1=40,x‾2=39,Sx2=5.8\overline{x}_1=41,S_{x1}=8.3,n_1=40,\overline{x}_2=39,S_{x2}=5.8, & n2=50n_2=50, perform a hypothesis test to test the claim that μ1>μ2\mu_1>\mu_2, assuming σ1=σ2\sigma_1=\sigma_2 for α=0.01\alpha=0.01.

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 = 41 , S x 1 = 8.3 , n 1 = 40 , x ‾ 2 = 39 , S x 2 = 5.8 \overline{x}_1=41,S_{x1}=8.3,n_1=40,\overline{x}_2=39,S_{x2}=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 , assuming σ 1 = σ 2 \sigma_1=\sigma_2 for α = 0.01 \alpha=0.01 .

Welcome back. For this problem, we have information on two samples, their mean, standard deviation, and size. We wanna perform a hypothesis test to test the claim that mu one is greater than mu two, assuming that the populations have the same standard deviation and variance. We have that alpha is equal to point zero one for this test. So let's start by writing our hypotheses. The null hypothesis is going to be that the two means are the same or that 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 let's pull out our t I 80 fours and run the test. We'll start by clicking on the stat button, scrolling to the right to the test menu, and then clicking on four. That's step one, all set. For step two, we wanna make sure that stats is selected, so make sure that your flashing cursor is over stats and then we'll enter in the statistics from this problem. So we have a sample mean of 41, standard deviation of 8.3, and size of 40 for that first sample. And for the second sample, the mean is 39, the standard deviation is 5.8, and the sample size is 50. Once that's all set, we need to make sure that the correct alternative hypothesis has been chosen. In this case, that should be that mu one is greater than mu two. So I'm gonna select the greater than mu two option. Also, the default for pooled is no, but in this case where I am assuming that the two variances are equal, I wanna select yes for pooled. So I'm gonna come over here and select yes for pooled. Then I wanna scroll down to calculate and click enter. That's step two, all set. And I'll see the results of the viewing window. The p value is approximately 0.09, and I wanna compare it to 0.01, which is the alpha value we've been given. I can see that the p value point zero nine is larger than point zero one alpha, which means that we fail to reject the null hypothesis and that there is not enough evidence to suggest that the alternative hypothesis is true, or to suggest that mu one is greater than mu two. Great work.

For x ‾ 1 = 41 , S x 1 = 8.3 , n 1 = 40 , x ‾ 2 = 39 , S x 2 = 5.8 \overline{x}_1=41,S_{x1}=8.3,n_1=40,\overline{x}_2=39,S_{x2}=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 , assuming σ 1 = σ 2 \sigma_1=\sigma_2 for α = 0.01 \alpha=0.01 .

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 .

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 .

10. Hypothesis Testing for Two Samples

Two Means - Unknown, Equal Variance

Statistics for Business

10. Hypothesis Testing for Two Samples

Two Means - Unknown, Equal Variance

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Multiple Choice

For ${\overline{x}}_{1} = 41, S_{x 1} = 8.3, n_{1} = 40, {\overline{x}}_{2} = 39, S_{x 2} = 5.8$ , & $n sub 2 equals 50$ , perform a hypothesis test to test the claim that $mu sub 1 is greater than mu sub 2$ , assuming $sigma sub 1 equals sigma sub 2$ for $alpha equals 0.01$ .

$P-value>\alpha$ ; 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

Step 1: Define the null and alternative hypotheses. Here, the claim is that $ \mu_1 > \mu_2 $, so set $ H_0: \mu_1 = \mu_2 $ and $ H_a: \mu_1 > \mu_2 $. This is a right-tailed test.

Step 2: Since $ \sigma_1 = \sigma_2 $ is assumed unknown but equal, use the pooled standard deviation to estimate the common population variance. Calculate the pooled variance $ S_p^2 $ using the formula: \[ S_p^2 = \frac{(n_1 - 1)S_{x1}^2 + (n_2 - 1)S_{x2}^2}{n_1 + n_2 - 2} \].

Step 3: Calculate the test statistic $ t $ using the formula: \[ t = \frac{\overline{x}_1 - \overline{x}_2}{S_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \], where $ S_p = \sqrt{S_p^2} $.

Step 4: Determine the degrees of freedom for the test, which is $ df = n_1 + n_2 - 2 $. Then, find the critical value from the t-distribution table for a right-tailed test at significance level $ \alpha = 0.01 $.

Step 5: Compare the calculated test statistic $ t $ with the critical value or calculate the p-value corresponding to $ t $. If the p-value is less than $ \alpha $, reject $ H_0 $; otherwise, fail to reject $ H_0 $.

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