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%.

A) Is the sample size large enough to justify the use of the Z formula?

B) Test if the proportion of the prevalence of HIV has changed (is equal to 7.5%). Use ? = 0.05.

C) Calculate the 95% two-sided confidence interval for p and make a conclusion about H0.

Carry probabilities to at least four decimal places for intermediate steps.

For extremely small probabilities, it is important to have one or two significant non-zero digits, for example, 0.000001 or 0.000034.

Round off your final answer to two decimal places.

*Show all intermediate steps*

Answer 1

a)By central limit theorem a large sample size can be approximated to normal. By 'large', we mean a sample size greater than 30. So, since the sample size is greater than 30, we can justify the use of the z formula. The number 30 is fixed by the statisticians and was found out even practically. So, there is no need to bother much aout the value 30.

B)Now, we need to test whether the proportion has changed from 7.5%.

i.e. H₀:p₀=0.075

Against H₁:p₀0.075

Here, the observed value=0.0994

The test statistic is given by

so, here, we have considered a total of 4706 samples. So, here, n= 4706

=468/4706=0.0994

Therefore, z=(.0994-0.075)/sqrt((0.0994*(1-0.0994))/4706)=5.5963.

And here the significance level =0.05

Therefore,Z_/2=1.96<5.5963

Since the value of the test statistic is greater than 1.96, we reject the null hypothesis and conclude that the proportion is not 7.5%.

C)

The Confidence interval is given by

Here, z=1.96, =0.004360.

where p=observed proportion=0.0994

So the interval is given by 0.09941.96*0.004360=(0.0908,0.1079).

Therefore we do not accept H₀ at 0.05% level.