For x‾1=44,Sx1=2.3,n1=50,x‾2=40,Sx2=3.1\overline{x}_1=44,S_{x1}=2.3,n_1=50,\overline{x}_2=40,S_{x2}=3.1, & n2=75n_2=75, test if there is evidence that μ1>μ2\mu_1>\mu_2 for α=0.01\alpha=0.01 using a hypothesis test.

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 .

For x ‾ 1 = 44 , S x 1 = 2.3 , n 1 = 50 , x ‾ 2 = 40 , S x 2 = 3.1 \overline{x}_1=44,S_{x1}=2.3,n_1=50,\overline{x}_2=40,S_{x2}=3.1 , & n 2 = 75 n_2=75 , test if there is evidence that μ 1 > μ 2 \mu_1>\mu_2 for α = 0.01 \alpha=0.01 using a hypothesis test.

Welcome back. For this problem, we have the mean, standard deviation, and size of two samples, and we'd like to test if there's evidence that mu one is greater than mu two for alpha equals 0.01 using a hypothesis test. We know that the null hypothesis is going to be that the two means are the same, mu one is equal to mu two. And the alternative hypothesis is what we're trying to test or see if there's evidence for, which in this case is that mu one is greater than mu two. Now that we have our hypotheses, we're ready to perform the test. We can skip step one because we've been given statistics instead of data and move straight to step two. Click on the stat button, two to the right to the test menu, and then click on four. That's step two, all set. We've been given stats, so we want to follow these instructions on the right, starting with scrolling to the right to stats, clicking enter, and then putting in the statistics we've been given. Our first sample mean is 44, the standard deviation is 2.3, and the size is 50. For that second sample, we have a mean of 40, a standard deviation of 3.1, and a sample size of 75. We wanna make sure that the correct alternative hypothesis has been chosen. In this case, we have that mu one is greater than mu two, so we want to make sure that greater than mu two has been selected. We want no for pooled, and then we can scroll down to calculate and click enter. We end up finishing step three with a p value of approximately 9.8 times 10 to the negative fourteenth, which we want to compare to 0.01, which is our alpha value. We can see that our p value is smaller than alpha, meaning that we reject our null hypothesis. That means that there is enough evidence to suggest that the alternative hypothesis is true, which in this case is that mu one is greater than mu two. Great work.

For x ‾ 1 = 44 , S x 1 = 2.3 , n 1 = 50 , x ‾ 2 = 40 , S x 2 = 3.1 \overline{x}_1=44,S_{x1}=2.3,n_1=50,\overline{x}_2=40,S_{x2}=3.1 , & n 2 = 75 n_2=75 , test if there is evidence that μ 1 > μ 2 \mu_1>\mu_2 for α = 0.01 \alpha=0.01 using a hypothesis test.

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 .

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, Unequal Variance

Statistics for Business

10. Hypothesis Testing for Two Samples

Two Means - Unknown, Unequal Variance

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Multiple Choice

For ${\overline{x}}_{1} = 44, S_{x 1} = 2.3, n_{1} = 50, {\overline{x}}_{2} = 40, S_{x 2} = 3.1$ , & $n sub 2 equals 75$ , test if there is evidence that $mu sub 1 is greater than mu sub 2$ for $alpha equals 0.01$ using a hypothesis test.

$P-value<\alpha$ ; 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 null hypothesis is

H_{0} : μ

₁ ≤ μ₂H_0: \mu_1 \leq \mu_2 and the alternative hypothesis is

H_{a} : μ

₁ > μ₂H_a: \mu_1 > \mu_2, indicating a one-tailed test.

Step 2: Calculate the test statistic for the difference between two means with independent samples. Use the formula:

z = \frac{\overline{x}_1 - \overline{x}_2}{\sqrt{\frac{S_{x1}^2}{n_1} + \frac{S_{x2}^2}{n_2}}}

Step 3: Substitute the given values into the formula:

\overline{x} 1 = 44, S_{x 1} = 2.3, n_{1} = 50, \overline{x} 2 = 40, S_{x 2} = 3.1, n_{2} = 75.

Step 4: Calculate the z-test statistic using the substituted values. This involves computing the numerator (difference of sample means) and the denominator (standard error of the difference).

Step 5: Determine the critical value or p-value corresponding to the significance level