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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. $$\begin{array}{cccc} \text { Kitchen } & \text { Master Bedroom } & \text { Kitchen } & \text { Master Bedroom } \\ 25.2 & 18.0 & 23.0 & 17.8 \\ 17.4 & 22.9 & 19.7 & 24.6 \\ 22.8 & 26.4 & 16.9 & 21.0 \\ 21.9 & 24.8 & 21.8 & \\ 19.7 & 26.9 & 23.6 & \end{array}$$ a. Develop a point estimate of the difference between the population mean remodeling costs for the two types of projects. b. Develop a \(90 \%\) confidence interval for the difference between the two population means.

Short Answer

Expert verified
The point estimate of the difference is -1.43, and the 90% confidence interval is (-4.033, 1.173).

Step by step solution

01

Organize Data

First, we need to separate the datasets:- Kitchen Remodeling Costs: \( [25.2, 23.0, 17.4, 22.8, 21.9, 19.7, 16.9, 21.8, 23.6] \)- Master Bedroom Remodeling Costs: \( [18.0, 17.8, 22.9, 26.4, 24.6, 21.0, 24.8, 26.9] \)
02

Calculate Sample Means

Calculate the sample mean for both datasets:For Kitchen Remodeling:\[ \bar{X}_1 = \frac{25.2 + 23.0 + 17.4 + 22.8 + 21.9 + 19.7 + 16.9 + 21.8 + 23.6}{9} = 21.47 \]For Master Bedroom Remodeling:\[ \bar{X}_2 = \frac{18.0 + 17.8 + 22.9 + 26.4 + 24.6 + 21.0 + 24.8 + 26.9}{8} = 22.9 \]
03

Calculate the Point Estimate

The point estimate of the difference between the two population means is the difference between the two sample means:\[ \bar{X}_1 - \bar{X}_2 = 21.47 - 22.9 = -1.43 \]
04

Calculate Sample Variances

Calculate the sample variance for each dataset:- For Kitchen Remodeling: \[ S_1^2 = \frac{1}{n_1-1} \sum (X_i - \bar{X}_1)^2 = \frac{51.468}{8} = 6.4335 \]- For Master Bedroom Remodeling: \[ S_2^2 = \frac{1}{n_2-1} \sum (Y_i - \bar{Y}_2)^2 = \frac{65.7225}{7} = 9.389 \]
05

Calculate Standard Errors

Next, calculate the standard error of the difference between means:- Using the formula for standard error: \[ SE = \sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}} = \sqrt{\frac{6.4335}{9} + \frac{9.389}{8}} = 1.486 \]
06

Calculate the Confidence Interval

Use the point estimate and standard error to calculate the confidence interval:- Determine the t-value for a 90% confidence interval with \(df = n_1 + n_2 - 2 = 15\): approx t-value is 1.753.- Confidence interval: \[ CI = \bar{X}_1 - \bar{X}_2 \pm t \times SE \] \[CI = -1.43 \pm 1.753 \times 1.486 \] \[CI = (-1.43 \pm 2.603)\] \[CI = (-4.033, 1.173)\]
07

Interpret the Results

The 90% confidence interval \((-4.033, 1.173)\) suggests there is no significant difference between the mean costs of kitchen and master bedroom remodeling projects, as zero is included in the interval.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Point Estimate
A point estimate helps us determine a specific numerical value that serves as an estimate for a population parameter. Essentially, it's a single value or statistic computed from the sample data, providing an approximation of an unknown parameter of the population. In our exercise, we calculated the point estimate of the difference between the mean remodeling costs of kitchens and master bedrooms.
The point estimate here is the difference between the sample means for the two groups. Specifically, in our example, this is calculated by subtracting the sample mean of Master Bedroom costs from the sample mean of Kitchen costs. It's important because it gives us a clear, focused measure of difference to work with, especially when proceeding to calculate confidence intervals.
Sample Mean
The sample mean is a fundamental concept in statistics that serves as a way to find the average value of a set of observations in a sample. The sample mean provides the most basic point estimate of the population mean, assuming that the sample is representative.
To calculate the sample mean, you sum all the data points in the sample and then divide by the number of data points. In our case, the sample mean for the Kitchen remodeling costs was calculated by summing all the provided costs and dividing by the number of kitchen projects (9). Similarly, the mean for Master Bedroom costs involves summing the costs and dividing by 8, the number of projects considered. These sample means serve as initial pieces in forming the point estimate.
Sample Variance
Sample variance is a useful statistic that gives us an idea of how data points in a sample are spread out around the sample mean. Essentially, it measures the variability or dispersion of the sample data.
To calculate the sample variance, you subtract the sample mean from each data point, square the result, sum these squared differences, and divide by the number of observations minus one. This results in the sample variances for Kitchen and Master Bedroom remodeling costs. In our example, these calculations provide us variance values of 6.4335 and 9.389 respectively. Appreciating the variance is crucial in understanding the spread of your data and in calculating the standard error.
Standard Error
The standard error is a valuable statistic that helps us understand the accuracy of a sample mean as an estimate of the population mean. It reflects how much the sample mean is expected to vary from sample to sample due to random chance.
In our context, the standard error of the difference between the two sample means is computed using the formula that incorporates both sample variances and the sample sizes. The formula considers the variance over their respective sample sizes, indicating how much variation is to be expected if new samples were taken. For our exercise, the calculated standard error was 1.486. This statistic is then used as a basis for constructing confidence intervals, providing a range in which we expect the actual population difference to lie.

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