Question

74 people enroll in a program.
Their mean weight loss was 24 lbs with sample standard deviation of 9.
A second group of 37 people didn't enroll and their weightloss was 15 lbs with a simple standard deviation of 6.
Construct a 98% confidence interval

Answer 1

To construct a 98% confidence interval for the difference in mean weight loss between the two groups, we can use a two-sample t-test approach because we have two independent samples with their means and standard deviations.

### Given Data:
- **Group 1 (Enrolled)**:
- Sample size (\( n_1 \)) = 74
- Mean weight loss (\( \bar{x}_1 \)) = 24 lbs
- Standard deviation (\( s_1 \)) = 9 lbs

- **Group 2 (Not Enrolled)**:
- Sample size (\( n_2 \)) = 37
- Mean weight loss (\( \bar{x}_2 \)) = 15 lbs
- Standard deviation (\( s_2 \)) = 6 lbs

### Step 1: Calculate the mean difference
The difference in mean weight loss between the two groups is:
\[
\bar{x}_1 - \bar{x}_2 = 24 - 15 = 9 \text{ lbs}
\]

### Step 2: Calculate the standard error of the difference in means
The standard error (\( SE \)) of the difference in means is calculated using the formula:
\[
SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}
\]
Calculating each term:
- \( s_1^2 = 9^2 = 81 \)
- \( s_2^2 = 6^2 = 36 \)

Now, substituting the values:
\[
SE = \sqrt{\frac{81}{74} + \frac{36}{37}}
= \sqrt{1.094595 + 0.972973}
= \sqrt{2.067568}
\approx 1.437
\]

### Step 3: Find the critical t-value
For a 98% confidence interval and degrees of freedom (\( df \)), we can approximate the degrees of freedom using:
\[
df = n_1 + n_2 - 2 = 74 + 37 - 2 = 109
\]
Using a t-distribution table or calculator, we can find the critical t-value for 98% confidence level with 109 degrees of freedom. The critical t-value for 98% confidence level (two-tailed) is approximately \( t_{0.01} \approx 2.626 \).

### Step 4: Calculate the margin of error
The margin of error (ME) is calculated using:
\[
ME = t \times SE
\]
Where \( t \) is the critical value. Substituting the values:
\[
ME = 2.626 \times 1.437 \approx 3.77
\]

### Step 5: Construct the confidence interval
Finally, the confidence interval for the difference in means is given by:
\[
(\bar{x}_1 - \bar{x}_2) \pm ME = 9 \pm 3.77
\]
Calculating the limits:
- Lower limit: \( 9 - 3.77 \approx 5.23 \)
- Upper limit: \( 9 + 3.77 \approx 12.77 \)

### Conclusion
The 98% confidence interval for the difference in mean weight loss between the enrolled group and the non-enrolled group is approximately:
\[
(5.23 \text{ lbs}, 12.77 \text{ lbs})
\]