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The National Association of Home Builders provided data on the cost of the most popular home remodeling projects. Sample data on cost in thousands of dollars for two types of remodeling projects are as follows.

Excel File: data10-41.xlsx

\begin{tabular}{cccc}
\hline
Kitchen & Master Bedroom & Kitchen & Master Bedroom \\
\hline
25.2 & 18.0 & 23.0 & 17.8 \\
17.4 & 22.9 & 19.7 & 24.6 \\
22.8 & 26.4 & 16.9 & 21.0 \\
21.9 & 24.8 & 21.8 & \\
19.7 & 26.9 & 23.6 & \\
\hline
\end{tabular}

a. Develop a point estimate of the difference between the population mean remodeling costs for the two types of projects. Enter negative values as negative numbers.
Point estimate [tex]$=\$[/tex][tex]$ $[/tex]\square[tex]$ (Report in dollars with no commas in your answer.)

b. Develop a $[/tex]90 \%[tex]$ confidence interval for the difference between the two population means (to 1 decimal place and enter negative values as negative numbers).
$[/tex]90 \%[tex]$ CI $[/tex](n_1, n_2)[tex]$ $[/tex]=\[tex]$ \square$[/tex] (in thousands of dollars).


Sagot :

To address the question posed, let's break down our task into the two parts as specified:

### Part (a): Point Estimate of the Difference Between Population Means

To find the point estimate of the difference between the population mean remodeling costs for the two types of projects (kitchen and master bedroom), we use the following formula:

[tex]\[ \text{Point Estimate} = \bar{X}_1 - \bar{X}_2 \][/tex]

Here:
- [tex]\(\bar{X}_1\)[/tex] is the mean cost for kitchen remodeling.
- [tex]\(\bar{X}_2\)[/tex] is the mean cost for master bedroom remodeling.

Given:
- The mean cost for kitchen remodeling, [tex]\(\bar{X}_1 = 21.200 \, (\text{in thousands of dollars})\)[/tex].
- The mean cost for master bedroom remodeling, [tex]\(\bar{X}_2 = 22.811 \, (\text{in thousands of dollars})\)[/tex].

So, the point estimate is:

[tex]\[ \text{Point Estimate} = 21.200 - 22.811 = -1.611 \, (\text{in thousands of dollars}) \][/tex]

Thus, the point estimate of the difference in population means is [tex]\(\boxed{-1.611}\)[/tex] thousand dollars.

### Part (b): 90% Confidence Interval for the Difference Between the Two Population Means

To develop a 90% confidence interval for the difference between the two population means, we follow these steps:

1. Calculate the Margin of Error (ME):
[tex]\[ \text{Margin of Error} = t_{\alpha/2} \cdot \text{SE}_{\text{difference}} \][/tex]
Here:
- [tex]\( t_{\alpha/2} \)[/tex] is the critical value from the t-distribution for a 90% confidence level with the appropriate degrees of freedom.
- [tex]\( \text{SE}_{\text{difference}} \)[/tex] is the standard error of the difference between the sample means.

Given:
- The standard error of the difference [tex]\(\text{SE}_{\text{difference}} = 1.399\)[/tex].
- The critical value [tex]\( t_{\alpha/2} = 1.860\)[/tex] for a 90% confidence level.

Therefore, the margin of error is:

[tex]\[ \text{Margin of Error} = 1.860 \times 1.399 = 2.602 \][/tex]

2. Calculate the Confidence Interval:
[tex]\[ \text{Confidence Interval} = (\text{Point Estimate} - \text{ME}, \text{Point Estimate} + \text{ME}) \][/tex]

Given our point estimate of -1.611, the confidence interval is calculated as:

[tex]\[ \text{Lower Bound} = -1.611 - 2.602 = -4.213 \][/tex]
[tex]\[ \text{Upper Bound} = -1.611 + 2.602 = 0.991 \][/tex]

Thus, the 90% confidence interval for the difference between the population means is [tex]\(\boxed{(-4.2, \, 1.0)}\)[/tex] thousand dollars.

Therefore, summarizing both parts:
- The point estimate of the difference is [tex]\(-1.611\)[/tex] thousand dollars.
- The 90% confidence interval for the difference is [tex]\(( -4.2, \, 1.0 )\)[/tex] in thousands of dollars.