A random sample of 81 credit sales in a department store showed an average sale of $68.00. From past data, it is known that the standard deviation of the populationσ is $27.00. a. Determine the standard error of the mean. What is the 95% confidence interval for the population mean? b. Assume that the same sample average sale was obtained from a random sample of 100 credit sales. What is the 95% confidence interval for the population mean? c. What is the effect of a larger sample size on the margin of error (E)

(a) Standard error is $3

95% confidence interval for the population mean is ($62.03, $73.97)

(b) 95% confidence interval for the population mean is ($62.64, $73.36)

(c) A larger sample size decreased the margin of error (E)

Explanation:

(a) Standard error = standard deviation ÷ √n = $27 ÷ √81 = $27 ÷ 9 = $3

Confidence interval = mean + or - margin of error (E)

mean = $68

sd = $27

n = 81

df = n - 1 = 81 - 1 = 80

confidence level = 95%

t-value corresponding to 80 df and 95% confidence level is 1.990

E = t×sd/√n = 1.990×27/√81 = $5.97

Lower limit = mean - E = 68 - 5.97 = $62.03

Upper limit = mean + E = 68 + 5.97 = $73.97

95% confidence interval is ($62.03, $73.97)

(b) n = 100

df = n - 1 = 100 - 1 = 99

t-value corresponding to 99 df and 95% confidence level is 1.9843

E = t×sd/√n = 1.9843×27/√100 = $5.36

Lower limit = mean - E = 68 - 5.36 = $62.64

Upper limit = mean + E = 68 + 5.36 = $73.36

95% confidence interval is ($62.64, $73.36)

(c) A larger sample size decreased the margin of error because the relationship between sample size and margin of error is inverse in which increase in one quantity leads to a decrease in the other quantity.

n

explanation: