Question

The mean and the standard deviation of the sample of 100 bank customer waiting times are x⎯⎯x¯ = 5.33 and s = 2.008. 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 [ , ].

(Click to select)NoYes , interval is (Click to select) less more than 6.

Answer 1

Solution:

The confidence interval for a population Mean when is unknown is as follows :

The sample population is normally distributed with a mean .

A 100(1- ) percent confidence interval for is ,

Here,

S is the sample standard deviation,

  is the t point that gives right hand tail ares of   under the t curve having n-1 degrees of freedom ,

and n is the sample size

we have the following information :

Sample size of bank customer n= 100

sample mean of waiting times of bank customer x¯ = 5.33

standard deviation of waiting times of bank customer s = 2.008

degrees of freedom =n-1= 99

= the population of all possible bank customer waiting times using new system

the table of normal point   for 99 degrees of freedom as follows:

100(1-)%	1-
95%	0.95%	0.05	0.025	1.984

The 95% confidence interval for the mean is obtained below

=[4.931 , 5.728]

The 95% confidence interval for the population of all possible bank customers waiting times using the new system is  [4.931 , 5.728]

we are 95% confidence is less than 6 minutes because all values in the interval are below 6 minutes. that means the maximum mean of all possible bank customer waiting times using the new system is 5.728