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to test whether the mean time needed to mix a batch of material is the …

Question

to test whether the mean time needed to mix a batch of material is the same for three manufacturers, the iowa chemical company obtained the following data on the time (in minutes) needed to mix the material. (table with columns: machine 1, machine 2, machine 3; rows with times: 20, 12, 14; 24, 30, 14; 14, 31, 11; 11, 11, 16). questions include: calculate the sum of squares (treatments), sum of squares (error), mean square (treatments), mean square (error), test statistic, p - value interpretation, conclusion, 95% confidence interval for the difference in mixing times for two manufacturers, and interpretation of the interval.

Explanation:

Response

To solve this problem, we'll perform a one - way ANOVA to test if the mean mixing times differ among the three manufacturers.

Step 1: Organize the data

Let's denote the three manufacturers as Group 1, Group 2, and Group 3. The data is as follows:

  • Group 1: \(20, 29, 14, 22\)
  • Group 2: \(12, 30, 31, 31\)
  • Group 3: \(14, 14, 11, 16\)
Step 2: Calculate the sum of each group (\(T_j\))
  • For Group 1 (\(n_1 = 4\)):

\(T_1=20 + 29+14 + 22=85\)

  • For Group 2 (\(n_2 = 4\)):

\(T_2 = 12+30 + 31+31 = 104\)

  • For Group 3 (\(n_3 = 4\)):

\(T_3=14 + 14+11 + 16 = 55\)

Step 3: Calculate the grand total (\(G\)) and the total number of observations (\(N\))

The total number of observations \(N=n_1 + n_2 + n_3=4 + 4+4 = 12\)
The grand total \(G=T_1+T_2+T_3=85 + 104+55 = 244\)

Step 4: Calculate the sum of squares between groups (\(SSB\))

The formula for \(SSB\) is \(\sum_{j = 1}^{k}\frac{T_{j}^{2}}{n_{j}}-\frac{G^{2}}{N}\), where \(k = 3\) (number of groups)
\(\frac{T_{1}^{2}}{n_{1}}=\frac{85^{2}}{4}=\frac{7225}{4}=1806.25\)
\(\frac{T_{2}^{2}}{n_{2}}=\frac{104^{2}}{4}=\frac{10816}{4}=2704\)
\(\frac{T_{3}^{2}}{n_{3}}=\frac{55^{2}}{4}=\frac{3025}{4}=756.25\)
\(\sum_{j = 1}^{k}\frac{T_{j}^{2}}{n_{j}}=1806.25+2704 + 756.25=5266.5\)
\(\frac{G^{2}}{N}=\frac{244^{2}}{12}=\frac{59536}{12}\approx4961.333\)
\(SSB = 5266.5-4961.333 = 305.167\)

Step 5: Calculate the sum of squares within groups (\(SSW\))

First, we calculate the sum of squares for each group.

  • For Group 1:

\(\sum x_{1i}^{2}=20^{2}+29^{2}+14^{2}+22^{2}=400 + 841+196+484 = 1921\)
\(SS_{1}=\sum x_{1i}^{2}-\frac{T_{1}^{2}}{n_{1}}=1921 - 1806.25 = 114.75\)

  • For Group 2:

\(\sum x_{2i}^{2}=12^{2}+30^{2}+31^{2}+31^{2}=144+900 + 961+961 = 2966\)
\(SS_{2}=\sum x_{2i}^{2}-\frac{T_{2}^{2}}{n_{2}}=2966 - 2704 = 262\)

  • For Group 3:

\(\sum x_{3i}^{2}=14^{2}+14^{2}+11^{2}+16^{2}=196+196 + 121+256 = 769\)
\(SS_{3}=\sum x_{3i}^{2}-\frac{T_{3}^{2}}{n_{3}}=769 - 756.25 = 12.75\)

\(SSW=SS_{1}+SS_{2}+SS_{3}=114.75 + 262+12.75 = 389.5\)

Step 6: Calculate the degrees of freedom
  • Degrees of freedom between groups (\(df_{B}\)): \(df_{B}=k - 1=3 - 1 = 2\)
  • Degrees of freedom within groups (\(df_{W}\)): \(df_{W}=N - k=12 - 3 = 9\)
Step 7: Calculate the mean square between groups (\(MSB\)) and mean square within groups (\(MSW\))

\(MSB=\frac{SSB}{df_{B}}=\frac{305.167}{2}=152.5835\)
\(MSW=\frac{SSW}{df_{W}}=\frac{389.5}{9}\approx43.2778\)

Step 8: Calculate the F - statistic

\(F=\frac{MSB}{MSW}=\frac{152.5835}{43.2778}\approx3.526\)

Step 9: Find the critical value and make a decision

For \(\alpha = 0.05\), \(df_{1}=2\) and \(df_{2}=9\), the critical value \(F_{crit}\) from the F - table is approximately \(4.256\)

Since the calculated \(F = 3.526

Step 10: For part (b) - Calculate the 95% confidence interval for the difference between Manufacturer 1 and 3

The formula for the confidence interval for the difference between two means \(\mu_1-\mu_3\) is \((\bar{x}_1-\bar{x}_3)\pm t_{\alpha/2,df_{W}}\sqrt{MSW(\frac{1}{n_1}+\frac{1}{n_3})}\)

  • \(\bar{x}_1=\frac{T_1}{n_1}=\frac{85}{4} = 21.25\)
  • \(\bar{x}_3=\frac{T_3}{n_3}=\frac{55}{4}=13.75\)
  • \(\bar{x}_1-\bar{x}_3=21.25 - 13.75 = 7.5\)
  • \(t_{\alpha/2,df_{W}}=t_{0.025,9}=2.262\)
  • \(\sqrt{MSW(\frac{1}{n_1}+\frac{1}{n_3})}=\sqrt{43.2778(\frac{1}{4}+\frac{1}{4})}=\sqrt{43.2778\times\frac{1}{2}}=\sqrt{21.6389}\approx4.652\)

The confidence interval is \(7.5\pm2.262\times4.652\)
Lower limit: \(7.5-2.262\times4.652=7.5 - 10.52= - 3.02\)
Upper limit: \(7.5 + 2.262\times4.652=7.5…

Answer:

s:

  • Sum of Squares Between: \(305.17\) (rounded to two decimal places)
  • Sum of Squares Within: \(389.5\)
  • Mean Square Between: \(152.58\) (rounded to two decimal places)
  • Mean Square Within: \(43.28\) (rounded to two decimal places)
  • F - statistic: \(3.53\) (rounded to two decimal places)
  • Since \(F = 3.53<4.26\) (critical value), we conclude that we cannot conclude that the mean time is not the same for all manufacturers.
  • 95% confidence interval for \(\mu_1-\mu_3\): \((- 3.02,18.02)\) (rounded to two decimal places)