Question

A group of 71 people enrolled in a weight-loss program that involved adhering to a special diet and to a daily exercise program. After six months, their mean weight loss was 26 pounds, with a sample standard deviation of 10 pounds. A second group of 42 people went on the diet but didn't exercise. After six months, their mean weight loss was 15pounds, with a sample standard deviation of 8 pounds.

Construct a 95% confidence interval for the mean difference in weight losses. Let uv1 denote the mean weight loss of the group with daily exercise.

Answer 1

To construct a 95% confidence interval for the mean difference in weight losses between two groups (Group 1 with exercise and Group 2 without exercise), we follow these steps:

Let's denote:
- \( \bar{x}_1 = 26 \) (mean weight loss of the exercise group)
- \( s_1 = 10 \) (standard deviation of the exercise group)
- \( n_1 = 71 \) (sample size of the exercise group)

- \( \bar{x}_2 = 15 \) (mean weight loss of the non-exercise group)
- \( s_2 = 8 \) (standard deviation of the non-exercise group)
- \( n_2 = 42 \) (sample size of the non-exercise group)

We want to find the confidence interval for the difference in means \( \mu_1 - \mu_2 \).

The formula for the standard error (SE) of the difference in means is:

\[
SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}
\]

Calculating the standard error:

\[
SE = \sqrt{\frac{10^2}{71} + \frac{8^2}{42}} = \sqrt{\frac{100}{71} + \frac{64}{42}}
\]

Calculating each term:

\[
\frac{100}{71} \approx 1.40845
\]
\[
\frac{64}{42} \approx 1.52381
\]

Adding these together:

\[
SE = \sqrt{1.40845 + 1.52381} = \sqrt{2.93226} \approx 1.711
\]

Next, we find the mean difference:

\[
\bar{x}_1 - \bar{x}_2 = 26 - 15 = 11
\]

Now we will calculate the critical value for a 95% confidence interval using the t-distribution. Since the sample sizes are 71 and 42, we can use the formula to calculate the degrees of freedom using the Welch-Satterthwaite equation:

\[
df \approx \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1 - 1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2 - 1}}
\]

Calculating the numerator:

\[
\left( \frac{10^2}{71} + \frac{8^2}{42} \right)^2 = (1.40845 + 1.52381)^2 = (2.93226)^2 \approx 8.5904
\]

Calculating the denominator:

\[
\frac{\left( \frac{10^2}{71} \right)^2}{70} + \frac{\left( \frac{8^2}{42} \right)^2}{41} \approx \frac{(1.40845)^2}{70} + \frac{(1.52381)^2}{41}
\]
\[
= \frac{1.98463}{70} + \frac{2.32264}{41} \approx 0.02835 + 0.05694 \approx 0.08529
\]

Thus,

\[
df \approx \frac{8.5904}{0.08529} \approx 100.79 \text{ (round down to 100)}
\]

Now we can find the critical value \( t^* \) for \( df \approx 100 \) at a 95% confidence level. The critical \( t \) value for a two-tailed test is approximately \( t^* \approx 1.984 \) (you may look this up in a t-table).

Now we can construct the confidence interval:

\[
CI = \left( (\bar{x}_1 - \bar{x}_2) - t^* \cdot SE, (\bar{x}_1 - \bar{x}_2) + t^* \cdot SE \right)
\]

Calculating the margin of error:

\[
t^* \cdot SE = 1.984 \cdot 1.711 \approx 3.3976
\]

Finally, we can construct the confidence interval:

\[
CI = \left( 11 - 3.3976, 11 + 3.3976 \right) = \left( 7.6024, 14.3976 \right)
\]

So the 95% confidence interval for the mean difference in weight losses is approximately:

\[
\boxed{(7.60, 14.40)}
\]