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 1980’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 55% to 65%.

We are going to conduct an additional focus group before we launch the product. What should the sample size be if we want a 90% CI to be within 5% of the actual value?

a.260

b.259.78

c.33

d.268

e.112

Answer 1

Solution:

A nationwide survey gives us estimates for proportions as an interval from 55% to 65%. So, we will take an average proportion as (55+65)/2 = 60%.

So, we have p = 60% = 0.60, Margin of error = E = 5% = 0.05, confidence level = 90%

Z value for 90% is given as 1.6449 (by using z-table or excel)

Sample size formula is given as below:

n = p*q*(Z/E)^2

Where, q = 1 – p = 1 – 0.60 = 0.40

n = 0.60*0.40*(1.6449/0.05)^2

n = 259.7468

n = 260

Required sample size = 260

Answer: a.260