A body mass index (BMI) between 20 and 25 indicates a normal weight. In a random survey of 783 people from Group A and 780 people from Group B, it was found that 269 people from Group A and 294 people from Group B were normal weight. Construct a 90% confidence interval of the difference in the proportion of people from Group A and people from Group B who are normal weight. Then interpret the interval.

Note the subscripts
a
and
b
represent people from Group A and people from Group B, respectively. Therefore sample 1 is the people from Group A and sample 2 is the people from Group B.

Incorrect   <
p
a
−
p
b
<

Round each answer to 4 decimal places.

To construct a 90% confidence interval for the difference in the proportion of people with normal weight between Group A and Group B, we will use the following steps:

### Step 1: Define the Sample Proportions
Let:
- \( p_a \) represent the proportion of people from Group A who have a normal weight.
- \( p_b \) represent the proportion of people from Group B who have a normal weight.

The sample proportions are calculated as follows:
\[
\hat{p}_a = \frac{x_a}{n_a} = \frac{269}{783}
\]
\[
\hat{p}_b = \frac{x_b}{n_b} = \frac{294}{780}
\]

Where:
- \( x_a = 269 \): Number of people with normal weight in Group A.
- \( n_a = 783 \): Total number of people in Group A.
- \( x_b = 294 \): Number of people with normal weight in Group B.
- \( n_b = 780 \): Total number of people in Group B.

### Step 2: Calculate the Difference in Sample Proportions
\[
\hat{p}_a - \hat{p}_b = \frac{269}{783} - \frac{294}{780}
\]

### Step 3: Calculate the Standard Error
The standard error for the difference in proportions is given by:
\[
SE = \sqrt{\frac{\hat{p}_a(1 - \hat{p}_a)}{n_a} + \frac{\hat{p}_b(1 - \hat{p}_b)}{n_b}}
\]

### Step 4: Find the Critical Value for a 90% Confidence Interval
For a 90% confidence level, the critical value \( z \) corresponding to a two-tailed test is 1.645.

### Step 5: Calculate the Margin of Error
\[
ME = z \times SE
\]

### Step 6: Construct the Confidence Interval
The confidence interval is given by:
\[
(\hat{p}_a - \hat{p}_b) - ME < p_a - p_b < (\hat{p}_a - \hat{p}_b) + ME
\]

Let's calculate each of these components.The 90% confidence interval for the difference in the proportion of people with normal weight between Group A and Group B is:

\[
-0.0733 < p_a - p_b < 0.0066
\]

### Interpretation:
This interval suggests that the proportion of people with normal weight in Group A could be up to 7.33% less than that in Group B or up to 0.66% more. Since this interval includes zero, it indicates there is no statistically significant difference at the 90% confidence level in the proportions of people with normal weight between the two groups.

Do you have any questions or would you like more details on any part of this process?

Here are some related questions you might find useful:
1. How would the confidence interval change with a 95% confidence level?
2. What is the interpretation if the confidence interval did not include zero?
3. How does the sample size affect the width of the confidence interval?
4. Can we use the same method for comparing more than two proportions?
5. What would happen to the confidence interval if one group had twice as many participants?

**Tip:** Confidence intervals can help you understand the potential range of differences between groups, especially when dealing with proportion data.

A body mass index (BMI) between 20 and 25 indicates a normal weight. In a random survey of 783 people from Group A and 780 people from Group B, it was found that 269 people from Group A and 294 people from Group B were normal weight. Construct a 90% confidence interval of the difference in the proportion of people from Group A and people from Group B who are normal weight. Then interpret the interval.

Learn how to construct a 90% confidence interval for the difference in proportions between two groups in a BMI survey. This problem explores key statistical concepts like sample proportions and standard error. Suitable for college students or advanced high school learners.

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