Question

(CO7) A state Department of Transportation claims that the mean wait time for various services at its different location is more than 6 minutes. A random sample of 16 services at different locations has a mean wait time of 9.5 minutes and a standard deviation of 7.6 minutes. At α=0.01, can the department’s claim be supported assuming the population is normally distributed?
@See text page 382

Yes, since p of 0.043 is greater than 0.01, fail to reject the null. Claim is null, so is supported

No, since p of 0.043 is greater than 0.01, reject the null. Claim is null, so is not supported

No, since p of 0.043 is greater than 0.01, fail to reject the null. Claim is alternative, so is not supported

Yes, since p of 0.043 is less than 0.09, reject the null. Claim is alternative, so is supported

Answer 1

Solution:

Given:

Sample size = n = 16

Sample mean =

Sample Standard Deviation = s = 7.6

Level of significance = α =0.01

Claim: the mean wait time for various services at its different location is more than 6 minutes.

We have to test if the department’s claim can be supported assuming the population is normally distributed.

Thus we use following steps:

Step 1) State H0 and H1:

Claim is alternative hypothesis H1.

Step 2) Find test statistic:

since sample size n is small and population standard deviation is unknown , we use t test statistic.

Step 3) Find p-value.

We need df = n - 1 = 16 - 1 = 15

Use following excel command to find p-value.

=T.DIST.RT( x , df )

=T.DIST.RT( 1.842 , 15 )

= 0.04267

=0.043

Thus p-value = 0.043

Step 4) Decision Rule:

Reject H0, if p-value < 0.01 level of significance, otherwise we fail to reject H0.

Since p-value = 0.043 > 0.01, we fail to reject H0.

Step 5) Conclusion:

Since we failed to reject H0, the department’s claim can not be supported

Thus correct choice is third option:

No, since p of 0.043 is greater than 0.01, fail to reject the null. Claim is alternative, so is not supported