Question

Exercises 8.41 to 8.45 refer to the data with analysis shown in the following computer output: \(\begin{array}{lrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } \\ \text { A } & 5 & 10.200 & 2.864 \\ \text { B } & 5 & 16.800 & 2.168 \\ \text { C } & 5 & 10.800 & 2.387\end{array}\) \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 2 & 133.20 & 66.60 & 10.74 & 0.002 \\ \text { Error } & 12 & 74.40 & 6.20 & & \\ \text { Total } & 14 & 207.60 & & & \end{array}\) Test for a difference in population means between groups \(A\) and \(C\). Show all details of the test.

Short answer

The t test statistic by given inputs is calculated. Then based on the calculated t value, the two-tailed p-value for our test is determined for 12 degrees of freedom. If this p-value is less than our significance level (commonly 0.05), we would conclude that there is a significant difference in the population means between groups A and C.

Step-by-step solution

Identify the necessary data

From the given diagrams, the relevant numbers are the Means and Standard Deviations (StDev) for groups A and C. A has Mean 10.200 with StDev 2.864 and C has Mean 10.800 with StDev 2.387. The number of elements (N) in each group is equal (5). The error degrees of freedom (DF) is also important which is 12.

Calculate the Pooled Standard Deviation

The pooled standard deviation (SDpooled) is given by the formula: \[\sqrt{\frac{(n_1-1)*StDev_1^2+(n_2-1)*StDev_2^2}{df}}\] Here \(n_1\) and \(n_2\) are the number of elements in each group (5 for each), \(StDev_1\) and \(StDev_2\) are the standard deviations for A and C (2.864 and 2.387 respectively), and df is the error degrees of freedom (12). Thus, \[SDpooled = \sqrt{\frac{(5-1)*2.864^2+(5-1)*2.387^2}{12}}\]

Calculate the test statistic

The t value is generally determined by using the formula \[t = \frac|{mean1-mean2}|{SDpooled*\sqrt{\frac{2}{n}}}\] Here, mean1 and mean2 are the means of groups A and C (10.200 and 10.800 respectively), \(SDpooled\) is the pooled standard deviation calculated in the previous step, and \(n\) denotes the number in each group (5). So, \[t = \frac|{10.200-10.800}|{SDpooled*\sqrt{\frac{2}{5}}}\]

Determine the p-value

Using a statistical table or calculator, find the two-tailed p-value for the t-statistic with 12 degrees of freedom. The two-tailed p-value indicates both the possibility that the mean of A could be greater than C and the mean of C could be greater than A.