Question

A company maintains a large fleet of automobiles. To check the average number of miles driven per month, a random sample of 100 cars is selected. The mean and standard deviation for the sample are 2,690 and 360, respectively. Records from previous years show that the average number of miles driven per month was 2,600. We wish to determine if the data provides sufficient evidence to support that there was a significant change in the average monthly mileage using a 5% level of significance. Step 4 of 5: If the test statistic was equal to 2.7, what would the p-value be equal to

Answer 1

Answer:

$df=n-1=100-1=99$  

Since is a two sided test the p value would be:  

$p_v =2*P(t_{(99)}>2.7)=0.0082$  

And since the p vaue is lower than the significance level we have enough evidence to reject the null hypothesis.

Step-by-step explanation:

Data given

$\bar X=2690$ represent the sample mean

$s=360$ represent the sample standard deviation

$n=100$ sample size  

$\mu_o =2600$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean of interest is different from 2600, the system of hypothesis would be:  

Null hypothesis:$\mu = 2600$  

Alternative hypothesis:$\mu \neq 2600$  

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

Calculate the statistic

The statistic for this case is given $t_{calc}=2.7$

P-value

The first step is calculate the degrees of freedom, on this case:  

$df=n-1=100-1=99$  

Since is a two sided test the p value would be:  

$p_v =2*P(t_{(99)}>2.7)=0.0082$  

And since the p vaue is lower than the significance level we have enough evidence to reject the null hypothesis.