Question

In a survey of adults with diabetes, the average body mass index (BMI) in a sample of 1931 women was 31.40 with a standard deviation of .3. The BMI in a sample of 1560 men was 30.60, with a standard deviation of .6.

(a) Construct a 99.5% confidence interval for the difference in the mean BMI between women and men with diabetes. Let denote the mean BMI for women with diabetes. Use the TI-84 calculator and round the answers to two decimal places.

Answer 1

To construct a 99.5% confidence interval for the difference in the mean BMI between women and men with diabetes, we will use the following formula for the confidence interval for the difference in means:

\[
\bar{X_1} - \bar{X_2} \pm z^* \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}
\]

Where:
- \(\bar{X_1} = 31.40\) is the mean BMI for women,
- \(\bar{X_2} = 30.60\) is the mean BMI for men,
- \(s_1 = 0.3\) is the standard deviation for women,
- \(s_2 = 0.6\) is the standard deviation for men,
- \(n_1 = 1931\) is the sample size for women,
- \(n_2 = 1560\) is the sample size for men,
- \(z^*\) is the z-score corresponding to the desired confidence level.

### Step 1: Find the z-score

For a 99.5% confidence interval, we look for the z-score that leaves 0.5% in the tails (0.25% in each tail). The critical value \(z^*\) can be found using the standard normal distribution table or a calculator:

\[
z^* \approx 2.576
\]

### Step 2: Calculate the standard error (SE)

Next, we calculate the standard error (SE):

\[
SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{0.3^2}{1931} + \frac{0.6^2}{1560}}
\]

Calculating the individual components:

\[
\frac{0.3^2}{1931} = \frac{0.09}{1931} \approx 0.0000465
\]
\[
\frac{0.6^2}{1560} = \frac{0.36}{1560} \approx 0.0002308
\]

Now summing those:

\[
SE = \sqrt{0.0000465 + 0.0002308} = \sqrt{0.0002773} \approx 0.01665
\]

### Step 3: Calculate the margin of error (ME)

Now, we calculate the margin of error:

\[
ME = z^* \cdot SE = 2.576 \cdot 0.01665 \approx 0.04288
\]

### Step 4: Calculate the confidence interval

Finally, we can find the confidence interval for the difference in means:

\[
\text{Difference} = \bar{X_1} - \bar{X_2} = 31.40 - 30.60 = 0.80
\]

Thus, the confidence interval is:

\[
0.80 \pm 0.04288
\]

Calculating the lower and upper bounds:

- Lower bound: \(0.80 - 0.04288 \approx 0.75712\)
- Upper bound: \(0.80 + 0.04288 \approx 0.84288\)

### Final Result

Rounding the results to two decimal places, we get:

\[
\text{99.5\% Confidence Interval} \approx (0.76, 0.84)
\]

So the 99.5% confidence interval for the difference in mean BMI between women and men with diabetes is \((0.76, 0.84)\).