Question

2. A local gym wants to determine whether there is a significant difference between the average
amount of time people spend exercising at the gym each month and the average amount of time
people spend exercising outdoors each month. The data they collected is shown in the table.
January February March April May June
Gym
6.2
4.1
3.4
2.2
0.8
1.1
Outdoors
1.5
0.9
1.7
3.2
4.3
4.2
Difference
4.7
3.2
1.7
-1
-3.5-3.1
(Gym - Outdoors)
How is the test statistic calculated? (Note that intermediate calculations have been rounded to 4
decimal places.)

Answer 1

Applying it's formula, it is found that the test statistic is of t = 0.24.

What is the formula for the test statistic?

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

• $\overline{x}$ is the sample mean.

• $\mu$ is the value tested at the null hypothesis.

• s is the standard deviation of the sample.

• n is the sample size.

In this problem, considering that we are testing if the means of two samples are different, the null hypothesis is $\mu = 0$.

The sample is given by: {4.7, 3.2, 1.7, -1, -3.5, -3.1}, hence:

$n = 6, \overline{x} = 0.3333, s = 3.3886$.

Then, the test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{0.3333 - 0}{\frac{3.3886}{\sqrt{6}}}$

$t = 0.24$

More can be learned about the test statistic of an hypothesis test at https://brainly.com/question/16313918