Question

A national organization sets out to investigate the change in prevalence of HIV since the last census in 2010. A total of 4,706 participants were interviewed and a total of 468 responses were confirmed to be HIV positive. Assume that the data from the census indicated that the prevalence of HIV in the particular population was 7.5%. Is the sample size large enough to justify the use of the Z formula?

Module Il P- b) 8 5 4706 -0.09945 0.90055 = 4249629 N=4060 = 4706 -0.099459 1.0 - 0.09945 = 0.90055 nee ='421.5 25... The Sáuple is large encrich to use the z formula, 16) n=4706 P=0.0994 8 = 0.9006 d=0.05 Ho = 7,5% = .075 Hat 7.5% = .025 Po = .025 %. = .925 Step 1 - Ho: P = Po Hailte Step 2 - Z statt Zatat = 0994-.0750 7007506.9.250) 1470.10 0244 - 0244 - 0244 = 6.4210 Zstat 1.0693): 4706 T. 00001473 0038 steps Zalat of 66.4210 = Pvalue 0. 00007 Step 4 Due to the pualue is 0.00007 which is less than the x of 0.05 in the results are statisticy significant and the Mo can be rejected with Lat 0.05 and p (= 0.00002), ic The change in the prevelence of HIV sunce 2010 with the Ho at 7.5% has been rejected. at the 20.05 with the p (=0.00007).

Answer 1

solution:

n=4706, x=468, p=0.075

np=4706*0.075=352.95

nq=n(1-p)=4706*(1-0.075)=0.4353

here np >5 and nq> 5

yes, sample size large enough to justify the use of the Z formula.

Bonus.

Now

Second Part: you have not mentioned in the question information about hypothesis, whether the test is one tailed or two tailed, as per your solution which is two tailed test, i am explaining.

The hypotheses are:

Ho : p=0.075 HQ :P + 0.075

The sample proportion is:

468 Pn4706 =770 = 0.0994

The test statistic is:

P-Po 0.0994 - 0.075 0.075(1-0.075) 4706 = 6.37 po(1-po)

P-value=P(|Z|>6.37)=2*P(Z<-6.37)=0.0000

(use z table)

Since p-value is 0.0000, reject null hypothesis.

Hence,

There is sufficient evidence to conclude that the proportion of the prevalence of HIV has changed from 7.5%.