We are tasked with calculating the t-test statistic for a one-sample t-test. The formula for the t-test statistic is:

$t = \frac{\bar{X} - \mu_0}{S.E.(\bar{X})}$

Where:

• $\bar{X}$ is the sample mean,
• $\mu_0$ is the population mean under the null hypothesis,
• $S.E.(\bar{X})$ is the standard error of the mean.

Step 1: Define the known values:

• Sample mean ($\bar{X}$) = 279,000
• Population mean under the null hypothesis ($\mu_0$) = 250,000
• Sample standard deviation (s) = 32,000
• Sample size (n) = 12

Step 2: Calculate the standard error of the mean ($S.E.(\bar{X})$):

The formula for the standard error of the mean is:

$S.E.(\bar{X}) = \frac{s}{\sqrt{n}}$

Substitute the known values:

$S.E.(\bar{X}) = \frac{32,000}{\sqrt{12}} \approx \frac{32,000}{3.464} \approx 9,236.353$

Step 3: Calculate the t-test statistic:

Now, substitute the values into the formula for the t-test statistic:

$t = \frac{279,000 - 250,000}{9,236.353} = \frac{29,000}{9,236.353} \approx 3.140$

Thus, the t-test statistic is approximately 3.140.

Would you like further details on how to interpret this result, or any additional explanation?

Here are five related questions:

1. How do you interpret the t-test statistic in terms of the null hypothesis?
2. What is the critical value of t for a significance level of 0.05 with 11 degrees of freedom?
3. How does increasing the sample size affect the standard error and the t-test statistic?
4. What assumptions are made when performing a t-test on this data?
5. How would the result change if the standard deviation were smaller?

Tip: A higher t-test statistic indicates stronger evidence against the null hypothesis.