A marketing researcher for a phone company surveys 300300 people and finds that the proportion of clients who are likely to switch providers when their contract expires is 0.160.16. a) What is the standard deviation of the sampling distribution of the proportion? b) If she wants to reduce the standard deviation by half, how large of a sample would she need?
Accepted Solution
A:
Answer:a) The standard deviation of the sampling distribution of the proportion is 0.021 = 2.1%.b) A sample of 1200 people.Step-by-step explanation:For each person, there are only two possible outcomes. Either they are likely to switch providers when their contract expires, or they are not. This means that we can solve this problem using the binomial probability distribution.Binomial probability distributionIt is the probability of exactly x sucesses on n repeated trials, with p probability.Has the following standard deviation of a sample proportion:[tex]\sqrt{V(X)} = \sqrt{\frac{p(1-p)}{n}}[/tex]In this problem, we have that:300 people were surveyed, so [tex]n = 300[/tex].The proportion of clients who are likely to switch providers when their contract expires is 0.16, so [tex]p = 0.16[/tex].a) What is the standard deviation of the sampling distribution of the proportion?[tex]\sqrt{V(X)} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.16*0.84}{300}} = 0.021[/tex]The standard deviation of the sampling distribution of the proportion is 0.021 = 2.1%.b) If she wants to reduce the standard deviation by half, how large of a sample would she need?The standard deviation of a sample is given by:[tex]s = \frac{\sqrt{V(X)}}{\sqrt{n}}[/tex]In this case, we have that:[tex]s = \frac{0.021}{\sqrt{n}}[/tex]So, the standard deviation of the sample is inverse proportional to the square root of the sample size. This means that to reduce the standard deviation by half, we need a sample size that is 4 times larger, that is, a sample of 4*300 = 1200 people.