Question

29) After elections were held, it was desired to estimate the proportion of voters who regretted that
they did not vote. How many voters must be sampled in order to estimate the true proportion to
within 2% (e.g., + 0.02) at the 90% confidence level? Assume that we believe this proportion lies
close to 30%.
A) n = 2017
B) n = 1421
C) n = 2401
D) Cannot determine because no estimate of p or q exists in this problem.
E) n = 1692

For the given binomial sample size and null-hypothesized value of p0, determine whether the sample size is large
enough to use the central limit theorem to conduct a test of the null hypothesis Ho: p = p0.
n = 700, p0 = 0.01
A) No
B) Yes

Answer 1

Let p be the true proportion of voters who regretted that they did not vote

Let be the sample proportion of voters in a sample of size n who regretted that they did not vote

The hypothesized value of p is

Using this we can calculate the standard error of proportion as

The significance level for 90% confidence interval is

The critical value of z for 90% confidence interval is

This can be written as

From the standard normal tables we get for z=1.64, P(Z<1.64) = 0.5+0.4495=0.9495 and for z=1.65 we get P(Z<1.65) = 0.5+0.4505=0.9505. Using linear interpolation we can get

.  

The 90% confidence interval is

But we want the true proportion to be within 2% (e.g., +- 0.02) at the 90% confidence level

That is we want a 90% confidence interval as

we get

The number of voters must be sampled in order to estimate the true proportion to within 2% (e.g., +- 0.02) at the 90% confidence level is n=1421

ans:B: n=1421

Using the central limit theorem, to approximate binomial distribution with a normal distribution, n has to be large enough that and have to be at least 5

For n=700 and

and

Since both are greater than 5, we can use central limit theorem and approximate the sampling distribution using a normal distribution

ans: Yes