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

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.

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{(\sigma^2_X)/(n_X)+(\sigma^2_Y)/(n_Y)}}\\`

`$$The calculated statistic is Z_c=\frac{[(7.79-7.64)-0]}{\sqrt{(1.06^2)/(80)+(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.