Question

1. A survey of first-time homebuyers found that the sample mean annual income was $46,000. Assume that the survey used a sample of 25 first-time home buyers and that the sample standard deviation was $1,100. Compute and explain a 95% confidence interval estimate of the population mean.

2. A telephone poll of 950 American adults asked: "where would you rather go in your spare time?" One response, by 275 adults, was "the Mall". Compute and explain a 99% confidence interval estimate of the proportion of all American adults who would respond "the Mall".

Answer 1

Solution :

1) The 95% confidence interval for population mean is given as follows:

Where, x̄ is sample mean, s is sample standard deviation and t(0.05/2, n-1) is critical t-value to construct 95% confidence interval.

We have, x̄ = $46000, s = $1100 and n = 25

Using t-table we get, t(0.05/2, 25-1) = 2.0639

Hence, 95% confidence interval for population mean is,

The 95% confidence interval to estimate the population is ($45545.942, $46454.058).

Explanation : We are 95% confident that true value of population mean of annual income of first-time home buyers would lie within the calculated 95% confidence interval.