Question

Exercise 2. A researcher compared the time spent on education in two countries, South Korea and the United States. The times spent on education by 65 students from South Korea had a mean of 48.5 hours per week. For 74 Students from the U.S., the mean was 47.3 hours per week. Assume the population standard deviation for South Korean students is 3.7 hours and it is 2.6 hours for the U.S. students. At 0.01 can the researcher support the claim that the mean time spent on education is less in the U.S.? a

For each exercise, complete the hypothesis test. Important! Please remember to include all 5 parts of a hypothesis test mentioned in the module summary. Assume that the samples are random, independent, and the populations are normally distributed.

Answer 1

Here we have: X1 48.5 2 47.3 1 3.7 2 2.6 n2 74 ni 65 α= 0.01 Hypothesis: Ho, H12 0 Ha, -H2> 0 (a known) Test statistics: z-distribution ( )-(, - pe2) σ N

(48.5-47.3)-(0) - 2.18 2.62 + 74 3.72 65 Critical value: Zo.01 2.326 Zcrit Za P-value p (t>2.18| =0.0146 Fail to reject the null hypothesis. Conclusion: Here we have not sufficient evidence to reject the null hypothesis. z-observed value is not greater than z-critical value. And p-value is greater than alpha value. We cannot support the claim that the mean spend on education is less in the U.S.

Non rejection region Rejection region Z-2.18 Žc 2.326