## Fowle Marketing Research Inc., bases charges to a client on the assumption that telephone surveys can be completed in a mean time of 15 minu

Fowle Marketing Research Inc., bases charges to a client on the assumption that telephone surveys can be completed in a mean time of 15 minutes or less. If a longer mean survey time is necessary, a premium rate is charged. A sample of 35 surveys provided the following survey times in minutes: 17;11;12;23;20;23;15;16;23;22;18;23;25;14;12;12;20;18;12;19;11;11;20;21;11;18;14;13;13;19; 16;10;22;18;23. Based upon past studies, the population standard deviation is assumed known with s = 4 minutes. Is the premium rate justified?

a. Formulate the null and alternative hypotheses for this application.

b. Compute the value of the test statistic.

c. What is the p-value?

d. At a = .01, what is your conclusion?

## Answers ( )

Answer:a) Null hypothesis:[tex]\mu \leq 15[/tex]

Alternative hypothesis:[tex]\mu > 15[/tex]

b) [tex]t=\frac{17-15}{\frac{4}{\sqrt{35}}}=2.958[/tex]

c) [tex]p_v =P(t_{34}>2.958)=0.0028[/tex]

d) If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 1% of signficance.

Step-by-step explanation:Data given and notation

The sample mean is given by:

[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

And the sample deviation is:

[tex] s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]

[tex]\bar X=17[/tex] represent the sample mean

[tex]s=4[/tex] represent the sample standard deviation

[tex]n=35[/tex] sample size

[tex]\mu_o =15[/tex] represent the value that we want to test

[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value for the test (variable of interest)

Part a State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less or equal than 15, the system of hypothesis would be:

Null hypothesis:[tex]\mu \leq 15[/tex]

Alternative hypothesis:[tex]\mu > 15[/tex]

Since we don’t know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)

t-test:“Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value”.Part b Calculate the statistic

We can replace in formula (1) the info given like this:

[tex]t=\frac{17-15}{\frac{4}{\sqrt{35}}}=2.958[/tex]

Part c P-value

The degrees of freedom are given by:

[tex] df = n-1= 35-1=34[/tex]

Since is a right tailed test the p value would be:

[tex]p_v =P(t_{34}>2.958)=0.0028[/tex]

Part d Conclusion

If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 1% of signficance.