Question

.1) How large a sample must one take to be 90% confident that the estimate is within 5% of the true

     proportion of women over 55 who are widows?  A recent study indicated that 29% of the 100 women

     over 55 in the study were widows.

8.  As a manufacturer of golf equipment, the Spalding Corporation wants to estimate the proportion of

     golfers who are left-handed.  (The company can use this information in planning for the number of

     right-handed and left-handed sets of golf clubs to make.) How many golfers must be surveyed if we

     want to be 99% confident that the sample proportion has a margin of error of 0.025?

            a.  Assume there is no estimate of a proportion available

            b.  Assume that we have an proportion estimate from a previous study that suggest that 15% of

                golfers are left-handed.

9.  Nielsen Media Research wants to estimate the mean amount of time (in minutes) that full-time college

     students spend watching television each weekday.  Find the sample size necessary to estimate the mean

     with a 15-minute margin of error.  Assume that a 95% confidence is desired, and the standard deviation

     is 112.2 minutes.

Answer 1

8)

Answer)

We need to use standard normal z table to answer this problem.

Critical value z for 99% confidence level is 2.58

Margin of error is given by critical value z *√{p*(1-p)/n}

N = sample size

Given margin of error = 0.025

A)

When no point of estimate p is given, we assume it to be 0.5

So, 0.025 = (2.58)*√0.5*(1-0.5)/√n

N = 2662.56 = 2663

B)

When p = 0.15

0.025 = (2.58)*√0.15*(1-0.15)/√n

N = 1357.9056 = 1358

9)

Critical value z for 95% confidence level is 1.96

Given margin of error = 15

Margin of error = z*(s.d/√n)

15 = 112.2/√n

N = 56