Answer:
Test statistics = 2.96
P-value = 0.35%
Yes, a premium rate is justified for this client
Explanation:
We are given that a sample of 35 surveys shows a sample mean of 17 minutes and standard deviation of 4 minutes.
Also, we have to test that whether telephone surveys can be completed in a mean time of 15 minutes or less or not.
Null Hypothesis,
:
<= 15 minutes {means that the telephone surveys can be completed in a mean time of 15 minutes or less}
Alternate Hypothesis,
:
> 15 minutes {means that the telephone surveys take more than mean time of 15 minutes to get completed}
The test statistics we will use here is;
T.S. =
~
where, X bar = sample mean = 17 minutes
s = sample standard deviation = 4 minutes
n = sample size = 35
So, Test statistics =
~
= 2.96
At 1% significance level, t table gives critical value of 2.441 at 34 degree of freedom. Since our test statistics is more than the critical value as 2.958 > 2.441 so we have sufficient evidence to reject null hypothesis as our test statistics will fall in the rejection region.
P-value is given by, P(
> 2.96) = 0.0035 or 0.35%
Here also, P-value is less than the significance level as 0.035% < 1% , so will reject null hypothesis.
Therefore, we conclude that the telephone surveys take more than mean time of 15 minutes to get completed and due to which a premium rate is charged and a premium rate is justified for this client.