To determine if the sample mean of 15 pushups differs significantly from the population mean of 12 pushups, we can use a one-sample t-test. The t-test compares the sample mean with the population mean to see if the difference between them is statistically significant.

Given data: - Sample size (\(n\)) = 25 - Sample mean (\(\bar{x}\)) = 15 pushups - Population mean (\(\mu\)) = 12 pushups - Standard deviation of the sample (\(s\)) = 9 pushups - Level of significance (\(\alpha\)) = 5% or 0.05

The formula to calculate the t-score is:

\[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \]

Substituting the given values:

\[ t = \frac{15 - 12}{\frac{9}{\sqrt{25}}} \]

\[ t = \frac{3}{\frac{9}{5}} \]

\[ t = \frac{3}{1.8} \]

\[ t = 1.67 \]

Now, with \(\alpha = 0.05\) and degrees of freedom (\(df\)) = \(n - 1 = 25 - 1 = 24\), we should check the t-distribution table to find the critical t-value for a two-tailed test because we are checking for any significant difference (not just an increase or a decrease). At the 5% significance level, the critical t-value for 24 degrees of freedom is approximately 2.064 (this value can slightly vary depending on the source).

The calculated t-value of 1.67 is less than the critical t-value of 2.064. This means we fail to reject the null hypothesis. In other words, the difference between the sample mean and the population mean is not statistically significant at the 5% level of significance. Therefore, based on this test, the average number of pushups (15) done by the pupils does not differ significantly from the population value of 12.