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

`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`

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 \\eq 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`