Question

Exercises 8.46 to 8.52 refer to the data with analysis shown in the following computer output: \(\begin{array}{lrrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } & \\ \text { A } & 6 & 86.833 & 5.231 & \\ \text { B } & 6 & 76.167 & 6.555 & \\ \text { C } & 6 & 80.000 & 9.230 & \\ \text { D } & 6 & 69.333 & 6.154 & \\ \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 3 & 962.8 & 320.9 & 6.64 & 0.003 \\ \text { Error } & 20 & 967.0 & 48.3 & & \\ \text { Total } & 23 & 1929.8 & & & \end{array}\) Find a \(95 \%\) confidence interval for the difference in the means of populations \(\mathrm{C}\) and \(\mathrm{D}\).

Short answer

Based on the computations and provided data, the 95% confidence interval for the difference in the means of populations C and D is \((2.98, 18.36)\).

Step-by-step solution

Identify relevant data

Firstly, recognize and note the required data for populations C and D. The mean for population C is 80.0, and for D it's 69.33. The standard deviation for population C is 9.23 and for D it's 6.15. The sample size (denoted by N) for both populations is 6.

Compute the standard error of the difference

Calculate the standard error of the difference denoted by \(S_{CD}\). This is computed by \(\sqrt{(s_C^2/n_C) + (s_D^2/n_D)}\) where \(s_C = 9.23\), \(s_D = 6.15\), \(n_C = 6\), and \(n_D = 6\). This gives \(S_{CD} = \sqrt{(9.23^2/6) + (6.15^2/6)} = 3.92\).

Identify the level of significance

Having derived the standard error, recall the level of significance. The problem instructs to find a 95% confidence interval, implying a 0.05 level of significance (100% - 95%). This indicates that the quantile value is 1.96 representing a two-sided test. (This quantile value can be found in Z-tables or using common statistical software)

Define and calculate the confidence interval

Finally, solve for the confidence interval. The formula is given as \(\mu_C - \mu_D \pm z \cdot S_{CD}\) where \(\mu_C, \mu_D\) are the means for populations C and D and \(z\) is the quantile value representing the confidence level. Substituting the values in gives \(80 - 69.33 \pm 1.96 \cdot 3.92\), this equates to \(10.67 \pm 7.69\). Hence the confidence interval is \(2.98, 18.36\).