Question

The average length of time customers spent on waiting at ABC Store’s check-out counter before a new P.O.S. system was installed was 4.3 minutes. A sample of 40 customers randomly selected at the check-out counter after the new P.O.S. system was installed showed that the average waiting time was 3.5 minutes with a sample standard deviation of 2.8 minutes. We wish to determine if the average waiting time after the P.O.S. system was installed is less than 4.3 minutes.

1.   State the null and alternative hypotheses to be tested. (2 Points)

2.   Compute the test statistic. (2 Points)

3.   Determine the critical value for this test at the 0.05 level of significance. (2 Points)

4.   What do you conclude at the 0.05 level of significance? (2 Points)

5.   Construct a 95% confidence interval for the average waiting time after the new P.O.S. system was installed.

Answer 1

1)

H0: = 4.3

Ha: < 4.3

2)

Test statistics

t = - / ( S / sqrt(n) )

= 3.5 - 4.3 / (2.8 / sqrt(40) )

= -1.81

3)

At 0.05 level with 39 df ,

Critical value = -1.685

4)

Since test statistics < -1.685, we have sufficient evidence to reject H0.

We conclude at 0.05 level that we have enough evidence to support the claim .

5)

t critical value (Two tailed) at 0.05 level with 39 df = 2.023

95% confidence interval for is

- t * S / sqrt(n) < < + t * S / sqrt(n)

3.5 - 2.023 * 2.8 / sqrt(40) < < 3.5 + 2.023 * 2.8 / sqrt(40)

2.60 < < 4.40

95% CI is ( 2.60 , 4.40 )