Major television networks conducted a joint poll of viewers and asked them if they felt that beer and other alcoholic beverage commercials targeted teenagers and young adults (those under 2121 years old). The results of the survey are as follows. Network Advertising SurveyAge GroupNumber SurveyedNumber of "Yes" Responses30 or Younger

1000

474

Older than 30

1000

519 Calculate a 90%90% confidence interval for the difference in the proportions of those older than 3030 and those 3030 or younger that believe alcoholic beverage commercials targeted teenagers and young adults. Let the 3030 or younger age group be Population 1 and let the older than 3030 age group be Population 2. Write your answer using interval notation and round the interval endpoints to three decimal places.




We are tasked with calculating a 90% confidence interval for the difference in proportions of two populations:

- **Population 1 (30 or younger)**: 1000 surveyed, 474 said "Yes"
- **Population 2 (Older than 30)**: 1000 surveyed, 519 said "Yes"

The goal is to calculate the confidence interval for the difference in proportions \( p_1 - p_2 \).

### Step 1: Define Proportions

The sample proportions for the two populations are:
\[
\hat{p_1} = \frac{474}{1000} = 0.474
\]
\[
\hat{p_2} = \frac{519}{1000} = 0.519
\]

### Step 2: Difference in Proportions

The difference in the sample proportions is:
\[
\hat{p_1} - \hat{p_2} = 0.474 - 0.519 = -0.045
\]

### Step 3: Standard Error

The standard error \( SE \) for the difference in proportions is given by:
\[
SE = \sqrt{\frac{\hat{p_1}(1 - \hat{p_1})}{n_1} + \frac{\hat{p_2}(1 - \hat{p_2})}{n_2}}
\]
where:
- \( \hat{p_1} = 0.474 \), \( n_1 = 1000 \)
- \( \hat{p_2} = 0.519 \), \( n_2 = 1000 \)

Substitute these values into the formula for standard error:
\[
SE = \sqrt{\frac{0.474 \times (1 - 0.474)}{1000} + \frac{0.519 \times (1 - 0.519)}{1000}}
\]

### Step 4: Z-value for 90% Confidence Interval

For a 90% confidence level, the corresponding Z-value is approximately 1.645.

### Step 5: Confidence Interval Formula

The confidence interval for the difference in proportions is given by:
\[
(\hat{p_1} - \hat{p_2}) \pm Z \times SE
\]

Now, I'll calculate the standard error and use the Z-value to find the confidence interval.The 90% confidence interval for the difference in the proportions of those older than 30 and those 30 or younger who believe that alcoholic beverage commercials target teenagers and young adults is approximately:

\[
(-0.082, -0.008)
\]

This means we are 90% confident that the true difference in proportions lies between \(-0.082\) and \(-0.008\), indicating that the proportion of people 30 or younger is likely lower than those older than 30 by about 0.008 to 0.082.

Would you like more details or have any further questions?

Here are five related questions to explore:

1. How would the confidence interval change if the confidence level were increased to 95%?
2. What effect does sample size have on the width of the confidence interval?
3. How can we interpret a confidence interval that includes zero?
4. What assumptions must hold for the confidence interval to be valid?
5. How would the analysis change if the number of "Yes" responses was different for either group?

**Tip:** Confidence intervals narrow as sample sizes increase, meaning more precise estimates for larger samples.

Major television networks conducted a joint poll of viewers and asked them if they felt that beer and other alcoholic beverage commercials targeted teenagers and young adults (those under 21 years old). The results of the survey are as follows: Network Advertising Survey Age Group Number Surveyed Number of 'Yes' Responses. 30 or Younger: 1000 surveyed, 474 'Yes' responses. Older than 30: 1000 surveyed, 519 'Yes' responses. Calculate a 90% confidence interval for the difference in the proportions of those older than 30 and those 30 or younger that believe alcoholic beverage commercials targeted teenagers and young adults. Let the 30 or younger age group be Population 1 and let the older than 30 age group be Population 2. Write your answer using interval notation and round the interval endpoints to three decimal places.

Calculate a 90% confidence interval for the difference in proportions of two age groups (30 or younger vs. older than 30) regarding their belief that alcoholic beverage commercials target teenagers. The problem uses sample proportions, standard error, and the Z-value for confidence intervals. Suitable for high school or college-level statistics students.

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