Question

In recent years the mean number of miles driven per year by residents of a particular state was 11,568. A researcher wishes to test, at the 5% level of significance, whether the mean is different now. In a random sample of 62 drivers, the mean and sample standard deviation of the number of miles driven last year were 10,822 and 1,741, respectively. The rejection region for the relevant test is:

Answer 1

Answer:

The degrees of freedom are given by:

$df =n-1= 62 -1=61$

And we want to find a critical value on the t distribution with 61 degrees of freedom who accumulates 0.025 of the area on each tail and the critical value would be $t_{\alpha/2}= \pm 2.00$

And the rejection region would be $|t_{calc}| >2.00$

Step-by-step explanation:

Data given and notation  

$\bar X=10822$ represent the sample mean  

$s=1741$ represent the sample standard deviation

$n=62$ sample size  

$\mu_o =11568$ represent the value that we want to test  

$\alpha=0.05$ 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)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean differs from 11568, the system of hypothesis would be:  

Null hypothesis:$\mu =11568$  

Alternative hypothesis:$\mu \neq 11568$  

The degrees of freedom are given by:

$df =n-1= 62 -1=61$

And we want to find a critical value on the t distribution with 61 degrees of freedom who accumulates 0.025 of the area on each tail and the critical value would be $t_{\alpha/2}= \pm 2.00$

And the rejection region would be $|t_{calc}| >2.00$