Answer:
(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.
