Question

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?

Answer 1

Answer:

a) Null hypothesis:$\mu \leq 15$  

Alternative hypothesis:$\mu > 15$  

b) $t=\frac{17-15}{\frac{4}{\sqrt{35}}}=2.958$      

c) $p_v =P(t_{34}>2.958)=0.0028$    

d) If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v$ 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:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And the sample deviation is:

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=17$ represent the sample mean    

$s=4$ represent the sample standard deviation

$n=35$ sample size  

$\mu_o =15$ represent the value that we want to test  

$\alpha=0.01$ represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

$p_v$ 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:$\mu \leq 15$  

Alternative hypothesis:$\mu > 15$  

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:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (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:  

$t=\frac{17-15}{\frac{4}{\sqrt{35}}}=2.958$  

Part c P-value  

The degrees of freedom are given by:

$df = n-1= 35-1=34$

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

$p_v =P(t_{34}>2.958)=0.0028$  

Part d Conclusion  

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v$ 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.