The Eastern and Western Major League Soccer conferences have a new Reserve Division that allows new players to develop their skills. Data for a randomly picked date showed the following annual goals for six different teams in each division.
\begin{tabular}{|r|r|}
\hline Eastern & Western \\
\hline 9 & 9 \\
\hline 3 & 8 \\
\hline 4 & 7 \\
\hline 3 & 6 \\
\hline 4 & 5 \\
\hline 4 & 3 \\
\hline
\end{tabular}
Does the data show there is a difference in the annual goals for the eastern and western divisions? Test the claim at the 0.05 significance level.
a) The null and alternative hypothesis would be:
\[
\begin{array}{l}
H_{0}: \mu_{E}=\mu_{W} \\
H_{1}: \mu_{E}>\mu_{W} \\
H_{0}: \mu_{E}=\mu_{W} \\
H_{1}: \mu_{E}<\mu_{W}
\end{array}
\]
\[
\begin{aligned}
H_{0}: \mu_{E} & =\mu_{W} \\
H_{1} & : \mu_{E} \neq \mu_{W} \\
H_{0} & : p_{E}=p_{W} \\
H_{1} & : p_{E}>p_{W} \\
H_{0} & : p_{E}=p_{W} \\
H_{1}: p_{E} &
Solution
Step 1 ：The null and alternative hypothesis are given by: \(H_{0}: \mu_{E} = \mu_{W}\) and \(H_{1} : \mu_{E} \neq \mu_{W}\)
Step 2 ：Calculate the mean and standard deviation for each division. For the Eastern division: Mean (\(\mu_{E}\)) = \((9+3+4+3+4+4)/6 = 4.5\) and Standard Deviation (\(\sigma_{E}\)) = \(\sqrt{((9-4.5)^2+(3-4.5)^2+(4-4.5)^2+(3-4.5)^2+(4-4.5)^2+(4-4.5)^2)/5} = 2.19\)
Step 3 ：For the Western division: Mean (\(\mu_{W}\)) = \((9+8+7+6+5+3)/6 = 6.33\) and Standard Deviation (\(\sigma_{W}\)) = \(\sqrt{((9-6.33)^2+(8-6.33)^2+(7-6.33)^2+(6-6.33)^2+(5-6.33)^2+(3-6.33)^2)/5} = 2.16\)
Step 4 ：Calculate the test statistic (t) as follows: \(t = (\mu_{E} - \mu_{W}) / \sqrt{(\sigma_{E}^2/n_{E}) + (\sigma_{W}^2/n_{W})}\), \(t = (4.5 - 6.33) / \sqrt{(2.19^2/6) + (2.16^2/6)}\), \(t = -1.83 / 0.91\), \(t = -2.01\)
Step 5 ：Find the p-value using a t-distribution table or calculator. For a two-tailed test with 10 degrees of freedom (6+6-2) and a t-score of -2.01, the p-value is approximately 0.0714.
Step 6 ：Since the p-value (0.0714) is greater than the significance level (0.05), we fail to reject the null hypothesis. \(\boxed{\text{Fail to Reject } H_{0}}\)
Step 7 ：There is not sufficient evidence to support the claim that there is a difference in the annual goals for the eastern and western divisions. \(\boxed{\text{Insufficient Evidence}}\)
From Solvely APP
Solution
Step 1 ：The null and alternative hypothesis are given by: \(H_{0}: \mu_{E} = \mu_{W}\) and \(H_{1} : \mu_{E} \neq \mu_{W}\)
Step 2 ：Calculate the mean and standard deviation for each division. For the Eastern division: Mean (\(\mu_{E}\)) = \((9+3+4+3+4+4)/6 = 4.5\) and Standard Deviation (\(\sigma_{E}\)) = \(\sqrt{((9-4.5)^2+(3-4.5)^2+(4-4.5)^2+(3-4.5)^2+(4-4.5)^2+(4-4.5)^2)/5} = 2.19\)
Step 3 ：For the Western division: Mean (\(\mu_{W}\)) = \((9+8+7+6+5+3)/6 = 6.33\) and Standard Deviation (\(\sigma_{W}\)) = \(\sqrt{((9-6.33)^2+(8-6.33)^2+(7-6.33)^2+(6-6.33)^2+(5-6.33)^2+(3-6.33)^2)/5} = 2.16\)
Step 4 ：Calculate the test statistic (t) as follows: \(t = (\mu_{E} - \mu_{W}) / \sqrt{(\sigma_{E}^2/n_{E}) + (\sigma_{W}^2/n_{W})}\), \(t = (4.5 - 6.33) / \sqrt{(2.19^2/6) + (2.16^2/6)}\), \(t = -1.83 / 0.91\), \(t = -2.01\)
Step 5 ：Find the p-value using a t-distribution table or calculator. For a two-tailed test with 10 degrees of freedom (6+6-2) and a t-score of -2.01, the p-value is approximately 0.0714.
Step 6 ：Since the p-value (0.0714) is greater than the significance level (0.05), we fail to reject the null hypothesis. \(\boxed{\text{Fail to Reject } H_{0}}\)
Step 7 ：There is not sufficient evidence to support the claim that there is a difference in the annual goals for the eastern and western divisions. \(\boxed{\text{Insufficient Evidence}}\)
