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Let's break down the given problem and solve it step by step to find the 90% confidence interval for the difference between the mean test scores of new hires from the current year (\( \mu_1 \)) and last year (\( \mu_2 \)).

### Given Information:
- **Current year (sample 1):**
  - Sample size (\( n_1 \)) = 50
  - Mean (\( \bar{x}_1 \)) = 113.3
  - Standard deviation (\( s_1 \)) = 12.8
- **Last year (sample 2):**
  - Sample size (\( n_2 \)) = 75
  - Mean (\( \bar{x}_2 \)) = 116.7
  - Standard deviation (\( s_2 \)) = 16.3
- Confidence level: 90%

### Step 1: Identify the formula for the confidence interval of the difference between two means
The formula for a confidence interval for the difference between two means is given by:

\[
(\bar{x}_1 - \bar{x}_2) \pm Z_{\alpha/2} \cdot \sqrt{\left(\frac{s_1^2}{n_1}\right) + \left(\frac{s_2^2}{n_2}\right)}
\]

Where:
- \( \bar{x}_1 - \bar{x}_2 \) is the difference between the sample means.
- \( Z_{\alpha/2} \) is the critical value from the standard normal distribution for a 90% confidence level.
- \( s_1^2 \) and \( s_2^2 \) are the sample variances of the two groups.
- \( n_1 \) and \( n_2 \) are the sample sizes.

### Step 2: Find the difference in means
\[
\bar{x}_1 - \bar{x}_2 = 113.3 - 116.7 = -3.4
\]

### Step 3: Find the critical value \( Z_{\alpha/2} \)
For a 90% confidence level, the critical value \( Z_{\alpha/2} \) is approximately **1.645** (from the Z-table).

### Step 4: Calculate the standard error
The standard error for the difference between two means is:

\[
SE = \sqrt{\left(\frac{s_1^2}{n_1}\right) + \left(\frac{s_2^2}{n_2}\right)} = \sqrt{\left(\frac{12.8^2}{50}\right) + \left(\frac{16.3^2}{75}\right)}
\]

First, calculate each component:
\[
\frac{12.8^2}{50} = \frac{163.84}{50} = 3.2768
\]
\[
\frac{16.3^2}{75} = \frac{265.69}{75} = 3.5425
\]

Now, calculate the standard error:
\[
SE = \sqrt{3.2768 + 3.5425} = \sqrt{6.8193} \approx 2.6114
\]

### Step 5: Calculate the confidence interval
Now, we can calculate the margin of error:

\[
\text{Margin of error} = Z_{\alpha/2} \cdot SE = 1.645 \cdot 2.6114 \approx 4.2948
\]

The confidence interval is:
\[
(\bar{x}_1 - \bar{x}_2) \pm \text{Margin of error} = -3.4 \pm 4.2948
\]

### Step 6: Calculate the lower and upper limits
- Lower limit: \( -3.4 - 4.2948 = -7.6948 \)
- Upper limit: \( -3.4 + 4.2948 = 0.8948 \)

### Final Answer:
- **Lower limit**: \( -7.69 \)
- **Upper limit**: \( 0.89 \)

Would you like further clarification or additional details?

Here are 5 related questions you can explore:
1. How is the standard error affected by sample size and variance?
2. Why is the critical value different for various confidence levels?
3. What would happen to the confidence interval if the sample sizes were smaller?
4. How does the confidence interval interpretation change with a different confidence level (e.g., 95%)?
5. What assumptions must be satisfied for this confidence interval to be valid?

**Tip:** Always check whether the sample sizes are large enough to apply normal approximation for confidence intervals. For smaller samples, a t-distribution might be required.

The human resources department of a consulting firm gives a standard creativity test to a randomly selected group of new hires every year. This year, 50 new hires took the test and scored a mean of 113.3 points with a standard deviation of 12.8. Last year, 75 new hires took the test and scored a mean of 116.7 points with a standard deviation of 16.3. Assume that the population standard deviations of the test scores of all new hires in the current year and the test scores of all new hires last year can be estimated by the sample standard deviations, as the samples used were quite large. Construct a 90% confidence interval for the difference between the mean test score of new hires from the current year and the mean test score of new hires from last year. Then find the lower limit and upper limit of the 90% confidence interval.

Learn how to construct a 90% confidence interval for the difference in mean test scores between two groups of new hires. This problem involves using the standard error formula and Z-scores to calculate the lower and upper limits. Data includes sample sizes, means, and standard deviations from a consulting firm's test results.

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