Question

21. The financial aid director of a large college reports that  38% of the students enrolled are receiving some sort of financial aid. A researcher claims the percentage is lower and finds (from sampling) 29 out of 116 students are receiving some sort of financial aid. At alpha = 0.05,  decide whether the financial aid director's report that 38% of the students are receiving some sort of financial aid holds up.   

For question #21 (multiple choice), choose the alternative hypothesis.

For #22, fill in the space provided with your calculated value for p hat.

For question #23, fill in the space provided with your calculated z-value from the information provided.

For #24, answer True or False with regard to rejecting the null hypothesis.

Answer 1

The following information is provided: The sample size is N = 116, the number of favorable cases is X 29, and the sample proportion is p = x = 0.25, and the significance level is a = 0.05 = 29 116 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho:p=0.38 Ha:p < 0.38 This corresponds to a left-tailed test, for which a z-test for one population proportion needs to be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a left-tailed test is Zc = -1.64. The rejection region for this left-tailed test is R = {z: z < -1.64} (3) Test Statistics The z-statistic is computed as follows: 2= p – po V po(1 – po)/n 0.25 – 0.38 (0.38(1 – 0.38)/116 2.885 (4) Decision about the null hypothesis -2.885 < Zc = -1.64, it is then concluded that the Since it is observed that z = null hypothesis is rejected. 0.002 Using the P-value approach: The p-value is p 0.002, and since p 0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population proportion p is less than po, at the a = 0.05 significance level.

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