Question

A simple random sample of 60 items resulted in a sample mean of 71. The population standard deviation is 13.

a. Compute the 95% confidence interval for the population mean (to 1 decimal). ( , )

b. Assume that the same sample mean was obtained from a sample of 120 items. Provide a 95% confidence interval for the population mean (to 2 decimals). ( , )

c. What is the effect of a larger sample size on the margin of error?

Answer 1

A simple random sample of 60 items resulted in a sample mean of 71. The population standard deviation is 13.

a. Compute the 95% confidence interval for the population mean (to 1 decimal). ( , )

Confidence interval for a population mean = Sample mean ±z * s /√n

Where,

Sample mean X = 71

s = standard deviation of population = 13

n = sample size = 60

Where z = 1.96 is a multiplier at 95% level of confidence

Therefore, Confidence intervals for population mean

= 71 - 1.96 * 13 /√60 to 71 + 1.96 * 13 /√60

= 71 - 3.2895 to 71 + 3.2895

= 67.7105 to 74.2895 or (67.7, 74.3)

Therefore from 95% confidence interval is between 67.7 and 74.3

b. Assume that the same sample mean was obtained from a sample of 120 items. Provide a 95% confidence interval for the population mean (to 2 decimals). ( , )

Confidence interval for a population mean = Sample mean ±z * s /√n

Where,

Sample mean X = 71

s = standard deviation of population = 13

n = sample size = 120

Where z = 1.96 is a multiplier at 95% level of confidence

Therefore, Confidence intervals for population mean

= 71 - 1.96 * 13 /√120 to 71 + 1.96 * 13 /√120

= 71 – 2.326 to 71 + 2.326

= 68.674 to 73.326 or (68.67, 73.33)

Therefore from 95% confidence interval is between 68.67 and 73.33

c. What is the effect of a larger sample size on the margin of error?

For larger sample size, the margin of error get reduced as the confidence interval get narrowed