A company decides that its employees should follow an exercise programme for 30 minutes each day, with the aim that they lose weight and increase productivity - CIE - A-Level Further Maths - Question 8 - 2011 - Paper 1

To find the sample mean loss:
$d = \frac{\sum d_i}{n} = \frac{(5.1 + 2.1 + 5.8 - 10.2 + 11.2 + 6.4 + 2.3 + 3.7)}{8} = 3.225$

To calculate the variance:
$s^2 = \frac{\sum (d_i - d)^2}{n - 1} = \frac{(5.1 - 3.225)^2 + (2.1 - 3.225)^2 + (5.8 - 3.225)^2 + (-10.2 - 3.225)^2 + (11.2 - 3.225)^2 + (6.4 - 3.225)^2 + (2.3 - 3.225)^2 + (3.7 - 3.225)^2}{7}$

Using the t-distribution for 7 degrees of freedom at 95% confidence level, the critical t-value is approximately $t_{0.025, 7} = 2.365$.
The confidence interval is calculated as:
$CI = d \pm t_{0.025, 7} \left( \frac{s}{\sqrt{n}} \right)$
This yields:
$CI = 3.225 \pm 2.365 \left( \frac{s}{\sqrt{8}} \right)$
Once values are plugged in, evaluate to find the specific interval.

Define the hypotheses for the significance test:

• Null Hypothesis ($H_0$): $\mu \leq 2.5$
• Alternative Hypothesis ($H_1$): $\mu > 2.5$

Using the sample mean found earlier, compute the test statistic:
$t = \frac{d - 2.5}{s / \sqrt{n}}$
Determine the p-value accordingly based on the computed t value and compare with the significance level.

Based on the comparison of the test statistic to the critical value from the t-table, draw a conclusion:

• If $t$ is greater than the critical value, reject $H_0$, indicating a significant reduction in weight. If not, fail to reject $H_0$.