Respuesta :
Answer:
Test statistics = 2.96
P-value = 0.35%
Yes, a premium rate is justified for this client
Step-by-step 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, [tex]H_0[/tex] : [tex]\mu[/tex] <= 15 minutes {means that the telephone surveys can be completed in a mean time of 15 minutes or less}
Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] > 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. = [tex]\frac{Xbar-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~  [tex]t_n_-_1[/tex]
where, X bar = sample mean = 17 minutes
        s = sample standard deviation = 4 minutes
        n = sample size = 35
So, Test statistics = [tex]\frac{17-15}{\frac{4}{\sqrt{35} } }[/tex] ~ [tex]t_3_4[/tex]
               = 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([tex]t_3_4[/tex] > 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.
