Question

The mean and the standard deviation of the sample of 100 bank customer waiting times are x⎯⎯ = 5.04 and s = 2.382. Calculate a t-based 95 percent confidence interval for µ, the mean of all possible bank customer waiting times using the new system.

Are we 95 percent confident that µ is less than 6 minutes?. (Choose the nearest degree of freedom for the given sample size. Round your answers to 3 decimal places.) The t-based 95 percent confidence interval is [ , ]. , interval is than 6.

Answer 1

Solution:

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± t*S/sqrt(n)

From given data, we have

Xbar = 5.04

S = 2.382

n = 100

df = n – 1 = 99

Confidence level = 95%

Critical t value = 1.9842

(by using t-table)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 5.04 ± 1.9842*2.382/sqrt(100)

Confidence interval = 5.04 ± 0.4726

Lower limit = 5.04 - 0.4726 = 4.567

Upper limit = 5.04 + 0.4726 = 5.513

Confidence interval = (4.567, 5.513)

The t-based 95 percent confidence interval is [4.567, 5.513].

The population mean µ is less than 6 minutes, because confidence interval is less than 6.