Question

QUESTION 2 The University of Auckland is considering the implementation of a week-long break during the fall semester. They wish to use a 90% confidence interval to estimate p, the proportion of students that favor this concept. Suppose 730 of the 879 surveyed students favor the concept. Hop=0.96 Hip+0.86 a) Compute the test statistic. Suppose you intend to evaluate the following hypothesis test on the proportion of interest and the resulting test statistics is -1.5. Hosp=0.86 Hip +0.86 b)Compute the P-value for this test. Suppose they surveyed 1.013 students to evaluate the following hypothesis test on the proportion of interest. Ho:p=0.86 Hip+0.86 c) What is the probability of type Il error if p=0.87?

Answer 1

0305 solution of x=730, na 079 sample psoportion hea = 730 879 USH, P7186 Ho P=.06 Ist we calculate gol. confidence Interval to estimate to P. 2 2.0 1.6450 c Å Joe x ²,2 (13) Z = 8305 t •83 OS X (1,8305) X 1.6450 &#9 = 8305 I 10 127 1.6450 r8305+ 10209 [.8096, 0514 ms

No. P= 186 vs HP 7,86 and test - Statistics value=z=+1,5) = 1.5 D kvalue = PIZ) > 1.50] a P{2>1.50] - 2x0668 la value 11336 ms Hol P= 186 HP & 186, he sample proportion Z-stata () = .0305 p - Pro 8305-86 = 10295 96X/y 879 .864.lu 079 12-stat 2.5137 PLF. 87 ZP 8 na 1013 Hoi p= 186 von Rob type and error=ß Mote that (1-B) = Prob of Rejecting Ho khere H: 82.87 <P13:26] > .06-.87 J-22x:13 .87X:13 = P11231-01 colo566 P[127.94643 = QP[22.94647 = 28.1726 13440 .656 1013 1013 > (SLD (ELD 15 seel