Question

A glass manufacturing company wanted to investigate the effect of zone 1 lower temperature​ (630 vs.​ 650) and zone 3 upper temperature​ (695 vs.​ 715) on the roller imprint of glass. Complete parts​ (a) through​ (e) below.

Zone 3 Upper   Zone 1 Lower   Roller Imprint
695 ---------------------630 ---------------51
695   -------------------630      -----------25
695   -------------------630 ---------- 52
695   -------------------630      -----------129
695   -------------------650      -----------24
695   -------------------650      -----------26
695   -------------------650      -----------48
695   -------------------650      -----------22
715   -------------------630      -----------98
715   -------------------630      -----------5
715   -------------------630      -----------25
715   -------------------630     ----------- 73
715   -------------------650      -----------76
715   -------------------650      -----------27
715   -------------------650      -----------1
715   -------------------650      -----------125

a. At the .05 level of​ significance, is there an interaction between zone 1 lower and zone 3​ upper?

Determine the test statistic.

Fstat=

Determine the p-value.

p-value=

b. At the .05 level of​ significance, is there an effect due to zone 1​ lower? First, determine the hypotheses.

Determine the test statistic.

Fstat=

Determine the p-value.

p-value=

c. At the .05 level of​ significance, is there an effect due to zone 3​ upper? First, determine the hypotheses.

Determine the test statistic.

Fstat=

Determine the p-value.

p-value=

d. Plot the roller imprint for each level of zone 1 lower for each level of zone 3 upper.

Plot the mean roller imprints for a zone 3 upper temperature of 695 with a dashed​ line, and the mean roller imprints for a zone 3 upper temperature of 715 with a solid line.

Answer 1

The test is performed in excel by using the following steps,

Step 1: Write the data values in excel. The screenshot is shown below,

Step 2: DATA > Data Analysis > ANOVA: Two Factor With Replication > OK.  The screenshot is shown below,

Step 3: Select Input Range: All the data values column, Rows per Sample: 4, Alpha = 0.05. The screenshot is shown below,

The result is obtained.  The screenshot is shown below,

a.

From the ANOVA table,

Fstat= 0.9722

p-value= 0.3436

Conclusion: Since the P-value is greater than the significance level = 0.05, the null hypothesis is rejected hence we can conclude that there is no interaction between zone 1 lower and zone 3​ upper temperature.

b.

From the ANOVA table,

Fstat= 0.4243

p-value= 0.5271

Conclusion: Since the P-value is greater than the significance level = 0.05, the null hypothesis is rejected hence we can conclude that there is no main effect due to zone 1​ lower temperature.

c.

From the ANOVA table,

Fstat= 0.1003

p-value= 0.7569

Conclusion: Since the P-value is greater than the significance level = 0.05, the null hypothesis is rejected hence we can conclude that there is no main effect due to zone 3​ upper temperature.

d.

The plot is obtained in excel in following steps,

Step 1: Write the mean value for each category. The screenshot is shown below,

Step 2:Select all the data values then INSERT > Recommended Charts > Scatters with Straight Line and Markers > Ok.