Question

Question 18 1 pts The Medical School Admission Test (MCAT) is a nationwide exam used to measure college students' readiness for medial school. Nationally, the mean score for all applicants is 30, and the standard deviation is 2.5. A certain university claims that their school's applicants have an average higher than 30. To test the claim, a random sample of 50 medical school applicants from the university is selected, with a mean MCAT score of 31. At a.05 significance level, is there evidence that this school's mean is higher than 30? No, because p= .0028 is less than the critical value Yes, because p= .0023 is less than .05. No. because z = 2.83 falls in the non-critical region Yes, because the p-value is less than 1.645 1 pts Question 19

Answer 1

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Here while calculating p-value for z-score probability we refer standard normal distribution table.

"enlivo Date Pogo page 0 Solution given that, Nationally the mean score for all applicants is 30 l=30 population standard deviation = 3=215 claim: Their schools applicants have an average higher than 30 Sample size=n= Sample mean = 8 - 31 level of significance. =&= 0.05 Nell and alternative hypothesis, Hoc el=30 No 4.4 730 Test statistici under Hou lie. 430. 50 a = 3- 30 BN 2.5/550 : 2-2,83 p-value & P(772083)= 1- P(Z < 2.88) = ) - 0.9977 -0.0023 [p-valu Value = 0.0023) Decision owe:- Reject Ho p-value ad = feired of significance Here p-value = 0.0023 We reject to at 5 % level of significance conclusion: There is evidence that this school's is higher than 30 Correct option is option B). Yes. because yuvalue = 0.0023 is less than 0.05 mean