Question

A recent study claims that business travelers spend an average of $41 per day on meals. A sample of 16 business travelers found that they had spent an average of $45 per day with a standard deviation of $3.65. If a = 0.05, what are the critical values?

Answer 1

Answer:

$df = n-1= 16-1 =15$

And the significance level is $\alpha=0.05$ and since we are conducting a bilateral test then the critical values are founded with the t distribution with 15 degrees of freedom and we got:

$t_{\alpha/2}= \pm 2.131$

Step-by-step explanation:

Information given

$\bar X=45$ represent the sample mean

$s=3.65$ represent the sample standard deviation

$n=16$ sample size  

$\mu_o =41$ represent the value that we want to test

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

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to determine the true mean is 41 per day, the system of hypothesis would be:  

Null hypothesis:$\mu = 41$  

Alternative hypothesis:$\mu \neq 41$  

We need to find the degrees of freedom first:

$df = n-1= 16-1 =15$

And the significance level is $\alpha=0.05$ and since we are conducting a bilateral test then the critical values are founded with the t distribution with 15 degrees of freedom and we got:

$t_{\alpha/2}= \pm 2.131$