Question

5) You now need to estimate with 90% confidence the population proportion of the amounts of financial aid that are at least $10,000. You randomly sample the population of financial aid awards that you are studying. You then record whether the amount of the award is at least $10,000. The results of that sampling process are shown in appendix two below. Using this information, construct the required confidence interval and explain its meaning. Appendix Two: (Award is at least $10,000?) (Y = yes, N = no) N Y N Y Y Y Y Y Y Y Y Y Y Y N人人人N人人人人人 NANANAN人人人人 N人人人人人N人人人人 人N人人人人人人 人人N人人人 人人人NN人人人N 人人人N人人人N人人人 人人NAL- N人人人人N人人人人N 人N人人人人人. 人人人人人人NNNN 人

Answer 1

From the data we find that 128 out of 176 recevied finacial aid of at least $10000.

90% confidence interval for the proportion of the amounts of finanical aid that are at least $10000 We are given Favorable events = S = 128 Sample size = n =176 128 176 = 0.7273 Sample proportion p = Confidence interval = 0.9 1 22 Finding the zstatistics a/2 = (1-0.9)/2 = 0.05 Using the ztable we find the zstat for a/2 = 0.05 zstat = 1.6449 p(1-P) 0.7273(1-0.7273) Standard Error = SE = = 0.0336 176 Margin of error = ME = zstat x SE = 1.6449 x 0.0336 = 0.0553 Confidence interval = p me = 0.7273 +0.0553 Therefore the 90% confidence interval for the proportion is (0.672, 0.7826) Hence we are 90% confident that the true population proportion who of the amounts of finanical aid that are at least $10000 lies between 0.672 and 0.7826