Question

We are planning on introducing a new internet device that should drastically reduce the amount of viruses on personal computers. We think the price should be $39.99, but are not sure on the percentage of people that would buy it. We do some research and find the following information; Studies from the 1930’s indicate that percentage should be between 30% and 40% Similar products were launched recently at a price of $4,000 and nobody bought it. A nationwide poll on this type of product and price was run earlier this year, with percentages running from 75% to 80%. We are going to conduct an additional focus group before we launch the product. What should the sample size be if we want a 95% CI to be within 5% of the actual value?

Answer 1

Answer:

The sample size required is 289.

Step-by-step explanation:

Let p be population proportion of people that would buy the product.

It is provided that the nationwide poll on this type of product and price was run earlier this year, with percentages running from 75% to 80%.

Assume that the sample proportion of people that would buy the product is, $\hat p=0.75$.

A 95% Confidence Interval is to be constructed with a margin of error of 5%.

We need to determine the sample size required for the 95% Confidence Interval to be within 5% of the actual value.

The formula to compute the margin of error for a (1 - α)% confidence interval of population proportion is:

$MOE=z_{\alpha/2}\times\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The critical value of z for 95% confidence interval is,

z = 1.96.

Compute the sample size required as follows:

$MOE=z_{\alpha/2}\times\sqrt{\frac{\hat p(1-\hat p)}{n}}$

      $n=[\frac{z_{\alpha/2}\ \sqrt{\hat p(1-\hat p)} }{MOE}]^{2}$

         $=[\frac{1.96\cdot \sqrt{0.75(1-0.75)} }{0.05}]^{2}\\\\=(16.9741)^{2}\\\\=288.12007081\\\\\approx 289$

Thus, the sample size required is 289.