Question

College national study finds that students buy coffee from a coffee shop on average 12 times a week, I believe it may be different for UML students. I collect data from a sample of 36 UML students and find that they buy coffee on average 8 times a week, with a standard deviation of 6 days. What is the T value for this data, and can you reject the null?

Answer 1

The t-value for this data is -4.

The p-value of the test is 0.0003 < 0.05, which means that the null hypothesis can be rejected.

Explanation:

College national study finds that students buy coffee from a coffee shop on average 12 times a week, I believe it may be different for UML students.

At the null hypothesis, we test if the mean is of 12, that is:

`H_0: \mu = 12`

At the alternative hypothesis, we test if the mean is different of 12, that is:

`H_1: \mu \\eq 12`

The test statistic is:

We have the standard deviation for the sample, so the t-distribution is used to solve this question.

`t = (X - \mu)/((s)/(√(n)))`

In which X is the sample mean,
`\mu` is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

12 is tested at the null hypothesis:

This means that
`\mu = 12`

I collect data from a sample of 36 UML students and find that they buy coffee on average 8 times a week, with a standard deviation of 6 days.

This means that
`n = 36, X = 8, s = 6`

Value of the test statistic:

`t = (X - \mu)/((s)/(√(n)))`

`t = (8 - 12)/((6)/(√(36)))`

`t = -4`

The t-value for this data is -4.

P-value of the test:

Considering a standard significance level of 0.05.

Test if the mean is different from a value, so two-tailed test, with 36 - 1 = 35 df and t = -4. Using a t-distribution calculator, the p-value is of 0.0003.

The p-value of the test is 0.0003 < 0.05, which means that the null hypothesis can be rejected.