Telemarketers Inc. hires over 2,200 hourly workers. The company wants to compute the average wage it pays these workers.

Telemarketers Inc. hires over 2,200 hourly workers. The company wants to compute the average wage it pays these workers. What is the minimum sample size needed to construct a 90% confidence interval for the average wage Telemarketer Inc. pays its hourly workers? Assume the standard deviation is $10.00 and the company wants a margin of error of $2.50.

Expert Solution

standard deviation = =$10.00

Margin of error = E = $2.50

At 90% confidence level

= 1 - 90%

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z 0.05 = 1.645

sample size = n = [Z/2 * / E] 2

n = ( 1.645 * 10.00/ 2.50)2

n =43.2964

Sample size = n =44

orchestra answered 2 years ago

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