A consumer's spending is widely believed to be a function of their income. To estimate this relationship, a university professor randomly selected 19 of his students and collected information on their spending (Y, in dollars) and income (X, in dollars) patterns in week 6 of the semester. Assuming a linear relationship between Y and X, the professor used the least-squares method and found that the Y intercept = 20.90 and the slope = 0.66. Also, the sum of squares total (SST) and the regression sum of squares (SSR) were equal to 65600.74 and 52831.23, respectively. What is the lower bound of a 95% confidence interval estimate of the population mean spending for a weekly income of $60. Assume the h Statistic equals 0.12.

	a.	
less than $41

	b.	
greater than $60.5

	c.	
$60.5

	d.	
greater than $41 but less than $60.5



To solve this problem, we'll break it down step by step. We're given the following information:

- Y-intercept ($b_0$) = 20.90
- Slope ($b_1$) = 0.66
- Sum of squares total (SST) = 65600.74
- Regression sum of squares (SSR) = 52831.23
- Weekly income ($X$) = $60
- $h$ Statistic = 0.12
- Confidence level = 95%

### Step 1: Calculate the predicted value of $Y$ for $X = 60$
The linear regression equation is:
\[
\hat{Y} = b_0 + b_1X
\]
Substituting the values:
\[
\hat{Y} = 20.90 + 0.66(60) = 20.90 + 39.6 = 60.5
\]
So, the predicted spending for an income of $60 is $60.5.

### Step 2: Calculate the Mean Squared Error (MSE)
First, calculate the error sum of squares (SSE):
\[
SSE = SST - SSR = 65600.74 - 52831.23 = 12769.51
\]
Now, calculate MSE:
\[
MSE = \frac{SSE}{n-2} = \frac{12769.51}{19-2} = \frac{12769.51}{17} \approx 751.15
\]

### Step 3: Determine the Standard Error of the Estimate ($SE_{\hat{Y}}$)
The standard error of the estimate for the prediction is given by:
\[
SE_{\hat{Y}} = \sqrt{MSE \cdot (1 + h)}
\]
Substitute the given values:
\[
SE_{\hat{Y}} = \sqrt{751.15 \cdot (1 + 0.12)} = \sqrt{751.15 \cdot 1.12} \approx \sqrt{841.288} \approx 29.01
\]

### Step 4: Calculate the t-statistic for the 95% confidence interval
For a 95% confidence level with 17 degrees of freedom ($n-2 = 17$), the critical value of $t$ is approximately 2.110 (this can be found in t-distribution tables).

### Step 5: Determine the margin of error
The margin of error is calculated as:
\[
\text{Margin of Error} = t \times SE_{\hat{Y}} = 2.110 \times 29.01 \approx 61.20
\]

### Step 6: Calculate the confidence interval
The lower bound of the confidence interval is:
\[
\hat{Y} - \text{Margin of Error} = 60.5 - 61.20 = -0.7
\]
However, a negative spending value does not make sense in this context. So, we should reconsider the interpretation; likely, the answer choices are not well-crafted. However, based on the most logical non-negative consideration, the interval's lower bound might suggest the range just below $41.

Given that context, **the most reasonable answer is:**

**a. less than $41.**

---

Would you like to explore the detailed calculations or have any questions? Here are some related topics to consider:

1. What is the importance of confidence intervals in regression analysis?
2. How is the t-distribution used in hypothesis testing?
3. How do changes in the slope affect predictions in linear regression?
4. What is the difference between SST, SSR, and SSE in regression analysis?
5. How does the value of $h$ influence the confidence interval?

**Tip:** When performing regression analysis, always check the assumptions of linearity, independence, homoscedasticity, and normality to ensure valid results.

Learn how to estimate the population mean spending using linear regression with a detailed example. Explore the use of least squares method, confidence intervals, and the t-distribution. Solve a practical problem involving income and spending data. Suitable for university-level students and researchers.

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