Question

A random sample of 36 elderly cellphone users revealed that, in a randomly selected month, they sent an average of 1200 text messages. Assuming that the number of text messages sent by elderly cellphone users in a month is normally distributed with a population standard deviation of 100 text messages, determine the following: • estimate, with 90% confidence, the true average number of text messages sent by the elderly per month. • what is the margin of error based on this sample? • if this study wanted the margin of error to be no greater than 15 text messages per month, what would the minimum sample size have needed to be?

Answer 1

Solution :

Given that,

Z/2 = Z_0.05 = 1.645

Margin of error = E = Z/2 * ( /n)

= 1.645 * (100 /  36 )

= 27.42

At 90% confidence interval estimate of the population mean is,

  ± E

1200  ± 27.42

(1172.58 , 1227.42)

Margin of error = E = Z/2 * ( /n)

= 1.645 * (100 /  36 )

= 27.42

sample size = n = [Z_/2* / E] ²

n = [ 1.645 * 100 / 15]²

n = 120.27

Sample size = n = 121