Question

All applicants to medical school are required to take an entrance exam. Historically, scores on the exam are known to be normally distributed with mean 80 and standard deviation 4. This year, we observe that a sample of 16 applicants obtains an average score of 78.67. Does this provide evidence that the average of all this year's applicants is less than 80? To answer this question, we do a hypothesis test using a significance level of 0.05.

Check all of the following statements that are correct (more than one alternative can be correct). Also, let ?μ be the mean of all this year's applicants.

Group of answer choices

a. The null hypothesis is ?0:?=H0:μ= 80

b. The alternative hypothesis includes an > sign

c. The value of the test statistic is z = -1.83

d. The p-value for this hypothesis test is 0.092

e. For this hypothesis test, we fail to reject ?0

Answer 1

Claim:  the average of all this year's applicants is less than 80

From the above claim the null hypothesis(H₀) and the alternative hypothesis(H₁) are as follows

H₀ : = 80

H₁ :   < 80

So the statement a) is correct

and statement b) is not correct.

The formula of test statistic is as follows:

..........( 1 )

Where = sample mean = 78.67 ,    = 80 ( from null hypothesis

n = sample size = 16 ,   = 4

Plugging these values in equation (1 ), we get :

The correct value of Z-test statistic = -1.33

So statement c) is not correct.

d) p-value = P(Z < -1.33) = 0.0918 = 0.092 (From Z-table)

So statement d) is correct

e) Decision rule:
1) If p-value < level of significance (alpha) then we reject null hypothesis
2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.
Here p value = 0.092 > 0.05 so we used 2nd rule.
That is we fail to reject null hypothesis.

So Statement e) is correct.