Question

A simple random sample with n=50provided a sample mean of 23.5 and a sample standard deviation of 4.3

a. Develop a 90% confidence interval for the population mean (to 1 decimal).

  ,  

b. Develop a 95% confidence interval for the population mean (to 1 decimal).

  ,  

c. Develop a 99% confidence interval for the population mean (to 1 decimal).

  ,  

d. What happens to the margin of error and the confidence interval as the confidence level is increased?

Answer 1

a)
sample mean, xbar = 23.5
sample standard deviation, s = 4.3
sample size, n = 50
degrees of freedom, n - 1 = 49

For 90% Confidence level, the t-value = 1.68

CI = (xbar - t*s/sqrt(n), xbar + t*s/sqrt(n))
CI = (23.5 - 1.68 * 4.3/sqrt(50) , 23.5 + 1.68 * 4.3/sqrt(50))
CI = (22.5 , 24.5)

b)
For 95% Confidence level, the t-value = 2.01

CI = (xbar - t*s/sqrt(n), xbar + t*s/sqrt(n))
CI = (23.5 - 2.01 * 4.3/sqrt(50) , 23.5 + 2.01 * 4.3/sqrt(50))
CI = (22.3 , 24.7)

c)
For 99% Confidence level, the t-value = 2.68

CI = (xbar - t*s/sqrt(n), xbar + t*s/sqrt(n))
CI = (23.5 - 2.68 * 4.3/sqrt(50) , 23.5 + 2.68 * 4.3/sqrt(50))
CI = (21.9 , 25.1)

d)
Increasing the value of CI, increases the value of ME and hence CI becomes wider.