In a current study, a random sample of 10 teachers from Texas and an independent random sample of 10 teachers from Indiana have been asked to report their annual income. The data obtained are as follows.

Annual income in dollars

Texas 48101, 22146, 39822, 31234, 25891, 28252, 38242, 42430, 35588, 34990

Indiana 49626, 44103, 38367, 28728, 29357, 57252, 39064, 38978, 28073, 36660

The population standard deviations for the annual incomes of teachers in Texas and in Indiana are estimated as $6200 and $6000, respectively. It is also known that both populations are approximately normally distributed. Construct a 90% confidence interval for the difference p₁₂ between the mean annual income of teachers from Texas (1) and the mean annual income of teachers from Indiana (4). Then find the lower limit and upper limit of the 90% confidence

interval

Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places. (If necessary, consult a list of

Lower limit:

Upper limit:

To construct a 90% confidence interval for the difference in the mean annual income between teachers from Texas and Indiana, we can follow these steps:

### Step 1: Calculate the sample means
Let's calculate the mean income for teachers in Texas and Indiana.

Given dataTo construct a 90% confidence interval for the difference in the mean annual income between teachers from Texas and Indiana, we can follow these steps:

### Step 1: Calculate the sample means
Let's calculate the mean income for teachers in Texas and Indiana.

Given 
- **Texas incomes:** 48101, 22146, 39822, 31234, 25891, 28252, 38242, 42430, 35588, 34990
- **Indiana incomes:** 49626, 44103, 38367, 28728, 29357, 57252, 39064, 38978, 28073, 36660

\[
\text{Mean for Texas } (\bar{X}_1) = \frac{1}{n_1} \sum X_{1i}
\]
\[
\text{Mean for Indiana } (\bar{X}_2) = \frac{1}{n_2} \sum X_{2i}
\]

Where $ n_1 = n_2 = 10 $.

### Step 2: Calculate the standard errors
Given the population standard deviations:
- $ \sigma_1 = 6200 $ (for Texas)
- $ \sigma_2 = 6000 $ (for Indiana)

The standard error for the difference between two independent means is calculated as:

\[
\text{Standard Error} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}
\]

### Step 3: Determine the critical value
For a 90% confidence interval, the critical value $ z_{\alpha/2} $ is required. Since it's a 90% confidence interval, $ \alpha = 0.10 $, and $ \alpha/2 = 0.05 $. The critical z-value for 0.05 in the standard normal distribution is approximately 1.645.

### Step 4: Construct the confidence interval
The confidence interval for the difference in means $ \mu_1 - \mu_2 $ is given by:

\[
(\bar{X}_1 - \bar{X}_2) \pm z_{\alpha/2} \times \text{Standard Error}
\]

### Step 5: Calculate the specific values
Let's compute the values.

I'll calculate these and provide the results.The 90% confidence interval for the difference between the mean annual incomes of teachers from Texas and Indiana is approximately:

- **Lower limit:** $-8839.37$
- **Upper limit:** $136.97$

This means we are 90% confident that the difference in mean annual incomes between Texas and Indiana teachers lies between $-8839.37$ and $136.97$ dollars.

Do you want more details, or have any questions?

Here are five related questions to explore further:
1. How does increasing the sample size affect the confidence interval?
2. What is the effect of changing the confidence level on the width of the interval?
3. How would the results change if the population standard deviations were different?
4. What is the importance of assuming normality in constructing confidence intervals?
5. How would the confidence interval change if the samples were not independent?

**Tip:** Remember that a confidence interval that includes zero suggests no significant difference between the two means at the given confidence level.

Learn how to construct a 90% confidence interval for the difference in mean annual income between teachers from Texas and Indiana. Includes step-by-step calculations using sample means, standard errors, and critical values. Suitable for advanced high school or college students.

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