Question

A football coach claims that players can increase their strength by taking a certain supplement. To test the theory, the coach randomly selects 9 athletes and gives them a strength test using a bench press. The results are listed below. Thirty days later, after regular training using the supplement, they are tested again. The new results are listed below. Test the claim that the supplement is effective in increasing the athletes' strength. Use α = 0.05. Verify anything that needs verification here. Athlete 1 2 3 4 5 6 7 8 9 Before 215 240 188 212 275 260 225 200 185 After 225 245 188 210 282 275 230 195 190

Answer 1

To test the claim that the supplement is effective in increasing the athletes' strength, a hypothesis test is conducted using a 5 percent level of significance. The null hypothesis is that the supplement does not increase strength, while the alternative hypothesis is that the supplement does increase strength. After calculating the test statistic and comparing it to the critical value, it is found that there is sufficient evidence to support the claim that the supplement is effective in increasing the athletes' strength.

Step-by-step explanation:

To test the claim that the supplement is effective in increasing the athletes' strength, we can conduct a hypothesis test. We will use a 5 percent level of significance (α = 0.05). The null hypothesis (H0) is that the supplement does not increase strength, while the alternative hypothesis (H1) is that the supplement does increase strength.

To conduct the test, we need to calculate the differences between the before and after strength test results. The differences are: 10, 5, 0, -2, 7, 15, 5, -5, 5. We find the sample mean of the differences to be 4.5 and the sample standard deviation to be 6.18.

Using these sample statistics, we can calculate the test statistic: t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size)). Under the null hypothesis, the hypothesized mean is 0 (no increase in strength). The calculated test statistic is t = 4.5 / (6.18 / sqrt(9)) = 2.74.

Next, we compare the test statistic to the critical value from the t-distribution at a 5 percent level of significance with 8 degrees of freedom. The critical value is approximately 2.31. Since the test statistic (2.74) is greater than the critical value (2.31), we reject the null hypothesis.

Therefore, based on the sample data, we have sufficient evidence to support the claim that the supplement is effective in increasing the athletes' strength.