• Historically, students in a statistics course receive an average score of 80 on the final exam with a standard deviation of 5. • One semester, 25 randomly chosen students are assigned a tutor who they meet with once a week. • Let X¯ be the average score on the final exam for those 25 students. • Assume that the standard deviation of final exam scores is completely unaffected by the tutoring, but that the tutoring could potentially increase exam scores (one sided test). Your null hypothesis is that tutoring has zero effect on final exam grades, and you are using a significance level of .05. (a) Under what values of X¯ would you reject the null hypothesis? (b) Suppose the null hypothesis is true. What is the probability of rejecting the null hypothesis? (c) Suppose, the null is false, and tutoring raises scores by 3 points for all students who receive it. What is the probability that X¯ will be far enough away from 80 for you to reject the null hypothesis? (d) What if tutoring raises scores by 1 point. What is the probability that X¯ will be far enough away from 80 for you to reject the null hypothesis? (e) How large would your sample have to be to have a 90% chance of rejecting the null hypothesis (assuming still that tutoring raises scores by 1 point).

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Answer:

Step-by-step explanation:

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The value of x where the researcher will reject the null hypothesis will be x > 81.96.

How to solve the hypothesis?

From the information given, it's a two tailed test with alpha of 0.05. We'll reject the null hypothesis when Z > 1.96.

Therefore, the value to reject the null hypothesis will be:

= (x - 80)/5/✓25 > 1.96

= x > 81.96

Assuming the null hypothesis is true, the probability of rejecting the null hypothesis will be:

= 1 - [(81.96 - 80)/(5/✓25)]

= 1 - (Z<1.96) - P[Z<-1.96]

= 1 - (0.975 - 0.025)

= 0.05

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