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Chapter 11: Problem 4

\( \begin{array}{ccccccccc} \text { Observation } & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} & \mathbf{7} & \mathbf{8} \\ \hline X_{i} & 19.4 & 18.3 & 22.1 & 20.7 & 19.2 & 11.8 & 20.1 & 18.6 \\ \hline Y_{i} & 19.8 & 16.8 & 21.1 & 22.0 & 21.5 & 18.7 & 15.0 & 23.9 \end{array} \) (a) Determine \(d_{i}=X_{i}-Y_{i}\) for each pair of data. (b) Compute \(\bar{d}\) and \(s_{d}\) (c) Test if \(\mu_{d} \neq 0\) at the \(\alpha=0.01\) level of significance. (d) Compute a \(99 \%\) confidence interval about the population mean difference \(\mu_{d}\).

Short Answer

Expert verified
The mean difference \( \bar{d} \) is -1.075. The null hypothesis is not rejected. The 99% confidence interval is (-5.149, 3.999).

Step by step solution

01

- Calculate the differences \(d_i\)

Compute the difference \(d_i = X_i - Y_i\) for each observation between \(X_i\) and \(Y_i\).
02

- Calculation

- For observation 1: \(d_1 = 19.4 - 19.8 = -0.4\)- For observation 2: \(d_2 = 18.3 - 16.8 = 1.5\)- For observation 3: \(d_3 = 22.1 - 21.1 = 1.0\)- For observation 4: \(d_4 = 20.7 - 22.0 = -1.3\)- For observation 5: \(d_5 = 19.2 - 21.5 = -2.3\)- For observation 6: \(d_6 = 11.8 - 18.7 = -6.9\)- For observation 7: \(d_7 = 20.1 - 15.0 = 5.1\)- For observation 8: \(d_8 = 18.6 - 23.9 = -5.3\)
03

- Compute the mean of differences \(\bar{d}\)

Add all the differences and divide by the number of observations:\(\bar{d} = \frac{\sum d_i}{n} = \frac{-0.4 + 1.5 + 1.0 -1.3 -2.3 -6.9 + 5.1 -5.3}{8} = \frac{-8.6}{8} = -1.075\)
04

- Compute the standard deviation of differences \(s_d\)

First, calculate the variance:\[s_d^2 = \frac{\sum(d_i - \bar{d})^2}{n-1}\]\(d_i - \bar{d}\) values:- \(-0.4 - (-1.075) = 0.675\)- \(1.5 - (-1.075) = 2.575\)- \(1.0 - (-1.075) = 2.075\)- \(-1.3 - (-1.075) = -0.225\)- \(-2.3 - (-1.075) = -1.225\)- \(-6.9 - (-1.075) = -5.825\)- \(5.1 - (-1.075) = 6.175\)- \(-5.3 - (-1.075) = -4.225\)Then calculate the squared values and sum them:- \(0.675^2 + 2.575^2 + 2.075^2 + (-0.225)^2 + (-1.225)^2 + (-5.825)^2 + 6.175^2 + (-4.225)^2 = 0.455625 + 6.630625 + 4.305625 + 0.050625 + 1.500625 + 33.927625 + 38.105625 + 17.840625 = 102.8175\)Now, divide by \(n-1 = 7\) and take the square root:\[s_d = \sqrt{\frac{102.8175}{7}} \approx 3.83\]
05

- Perform the t-test for \(\mu_d eq 0\)

Calculate the t-statistic:\[t = \frac{\bar{d} - 0}{s_d/\sqrt{n}} = \frac{-1.075}{3.83/\sqrt{8}} \approx -0.792\]Compare the t-statistic to the critical value for a two-tailed test with \(\alpha = 0.01\) and \(df = 7\). The critical t-value is approximately \( \pm3.499\). Since \(-0.792\) is within the range \([-3.499, 3.499]\), we do not reject the null hypothesis.
06

- Calculate the 99% confidence interval for \(\mu_d\)

Find the margin of error using the critical t-value and \(s_d\):\[\text{Margin of Error} = t_{\alpha/2, df} \times \frac{s_d}{\sqrt{n}} = 3.499 \times \frac{3.83}{\sqrt{8}} \approx 4.074\]Construct the confidence interval:\[\bar{d} \pm \text{Margin of Error} = -1.075 \pm 4.074 = (-5.149, 3.999)\]

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Key Concepts

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

mean difference
The mean difference, often denoted as \(\bar{d}\), is the average of the differences between paired observations in a data set. To compute the mean difference, subtract each paired value (Y_i) from its corresponding pair (X_i) to get the differences d_i. Sum all the differences and divide by the total number of observations (n). For the given data, the differences are computed as follows:
\(d_1 = 19.4 - 19.8 = -0.4\)
\(d_2 = 18.3 - 16.8 = 1.5\)
\(d_3 = 22.1 - 21.1 = 1.0\)
\(d_4 = 20.7 - 22.0 = -1.3\)
\(d_5 = 19.2 - 21.5 = -2.3\)
\(d_6 = 11.8 - 18.7 = -6.9\)
\(d_7 = 20.1 - 15.0 = 5.1\)
\(d_8 = 18.6 - 23.9 = -5.3\)
Summing these differences: \( -0.4 + 1.5 + 1.0 + -1.3 + -2.3 + -6.9 + 5.1 + -5.3 = -8.6\). Then dividing by the number of observations (8): \( \bar{d} = \frac{-8.6}{8} = -1.075\). The mean difference tells you how much the values in group X generally differ from those in group Y.
standard deviation of differences
The standard deviation of differences, denoted as \(s_d\), measures the variability of the differences between paired observations. Variability shows how spread out these differences are. To compute \(s_d\), first, get each difference \(d_i\) and subtract the mean difference \( \bar{d}\) from each. Squaring these results will handle any negative numbers. For the given data:
\( d_i - \bar{d} \):
\[ 0.675, \ 2.575, \ 2.075, \ -0.225, \ -1.225, \ -5.825, \ 6.175, \ -4.225 \]
Squared values:
\[ 0.455625, \ 6.630625, \ 4.305625, \ 0.050625, \ 1.500625, \ 33.927625, \ 38.105625, \ 17.840625 \]
Summing these squares: \( 102.8175\). Then divide this sum by \(n-1 = 7\): \[ s_d^2 = \frac{102.8175}{7} = 14.688\]. Taking the square root yields: \( s_d = \frac{102.8175}{7} = 3.83\). This gives the average distance of each \(d_i\) from the mean difference.
confidence interval
A confidence interval provides a range of values that likely contains the population mean difference \( \mu_d \). For a 99% confidence interval, we need the mean difference \( \bar{d} \), the standard deviation of the differences \( s_d \), and the critical value from the t-distribution (\

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