Question

A statistics instructor who teaches a lecture section of 160 student wants to determine weather the students have more difficulty with one-tailed hypothesis test or with two-tailed hypothesis test. On the next exam, 80 of stuents, choosen at random, get the version of the exam with a 10 point question that required a one-tailored test. the other 80 students get the question that is identical except that it required a two tailed-test. The one-tailed students average at 7.79 point, and the i standard deviation is 1.06 points. The two tailed students average 7.64 point, and their standard deviation is 1.31 point.
Can you conclude that the mean score μX on one-tailed hypothesis test questions is higher than the mean score μY on two-tailed hypothesis test questions? State the appropriate null and alternate hypotheses, and then compute the P-value. Check all that are true.

Answer 1

Answer:

There is not enough statistical evidence to state that the mean score on one-tailed hypothesis test questions is higher than the mean score on two-tailed hypothesis test questions.

Step-by-step explanation:

To solve this problem, we run a hypothesis test about the difference of population means.

$Sample mean \bar X: \bar X=7.79 \\Sample mean \bar Y: \bar Y=7.64 \\Population variance S^2_X: S^2_X=1.06^2 \\Population variance S^2_Y: S^2_Y=1.31^2 \\Sample size n_X=80 \\Sample size n_Y=80 \\Significance level \alpha=0.05 \\Z criticals values (for a 0.05 significance)\\(Left tail test) Z_{1-\alpha}=Z_{0.95}=-1.64485\\( Right tail test) Z_{\alpha}=Z_{0.05}=1.64485\\( Two-tailed test) Z_{1-\alpha/2}=Z_{0.975}=-1.95996 and Z_{\alpha/2}=Z_{0.025}=1.95996$

The appropriate hypothesis system for this situation is:

$H_0:\mu_X-\mu_Y=0\\H_a:\mu_X-\mu_Y > 0\\\\ Difference of means in the null hypothesis is:\\\mu_X-\mu_Y=M_0=0\\\\ The test statistic is Z=\frac{[( \bar X-\bar Y)-M_0]}{\sqrt{\frac{\sigma^2_X}{n_X}+\frac{\sigma^2_Y}{n_Y}}}$

$The calculated statistic is Z_c=\frac{[(7.79-7.64)-0]}{\sqrt{\frac{1.06^2}{80}+\frac{1.31^2}{80}}}=0.79616\\\\p-value = P(Z \geq Z_c)=0.42594$

Since, the calculated statistic $Z_c$ is less than critical $Z_{\alpha}$, the null hypothesis do not should be rejected. There is not enough statistical evidence to state that the mean score on one-tailed hypothesis test questions is higher than the mean score on two-tailed hypothesis test questions.