Question

The data table contains waiting times of customers at a​ bank, where customers enter a single waiting line that feeds three teller windows. Test the claim that the standard deviation of waiting times is less than 1.5 ​minutes, which is the standard deviation of waiting times at the same bank when separate waiting lines are used at each teller window. Use a significance level of 0.05. Complete parts​ (a) through​ (d) below.

A. Identify the null and alternative hypotheses for this test.

B. Identify the test statistic for this hypothesis test.

C. Identify the​ P-value for this hypothesis test.

D. Identify the conclusion for this hypothesis test.

Customer wait time (in minute):

8.6

7.3

6.2

6.5

6.4

6.6

6.6

6.4

7.9

7.6

6.4

8.9

11.2

6.1

8.9

7.8

7.6

6.4

7.1

6.5

6.7

6.7

7.2

6.2

7.2

7.1

6.4

7.8

7.6

6.3

4.2

6.4

7.8

7.1

6.5

6.3

7.9

6.8

7.4

7.4

10.7

6.4

6.5

6.3

7.6

7.3

6.1

7.8

7.9

7.4

5.3

8.8

6.6

6.2

7.1

6.3

7.5

7.3

7.4

6.3

Answer 1

From given sample data ,

sample size = n = 60

Sample standard deviation = s = 1.10645           { Using Excel function , =STDEV( select data )   }

Test the claim that the standard deviation of waiting times is less than 1.5 ​minutes.

Hypothesis :

   ( Claim )

Left tailed test.

Test statistic :

P-value :

P-value for this left tailed test is ,

P-value = P( < test statistic )

P-value = P( < 32.102 )

Using Excel function ,    =CHISQ.DIST( , df , 1 )  

df = n - 1 = 60 - 1 = 59

=CHISQ.DIST( 32.102 , 59 , 1 ) = 0.0016

P-value = 0.0016

Decision about null hypothesis :

Rule : Reject null hypothesis if p-value less than significance level .

Given significance level .= 0.05

It is observed that p-value ( 0.0016 ) is less than = 0.05

So reject null hypothesis.

Conclusion :

There is sufficient evidence to conclude that the standard deviation of waiting times is less than 1.5 ​minutes