Chapter 14, Problem 25CAP

1.

To determine

Complete the F table.

Identify whether the decision to retain or reject the null hypothesis for each hypothesis test

Answer to Problem 25CAP

The completed F table is,

Source of Variation	SS	df	MS	$F_{\text{obt}}$
Length	15	1	15	3.00
Technology	120	1	120	24.00
Length $\times$ Technology	144	1	144	28.80
Error	570	114	5
Total	849	117

The decision is to retain the null hypothesis for Length, the group means for main effect Length do not significantly vary in the population.

The decision is to reject the null hypothesis for Technology, the group means for main effect Technology do significantly vary in the population.

The decision is to reject the null hypothesis for interaction effect; the means of Lengthdo significantly vary by Technology or the combinations of these factors.

Explanation of Solution

Calculation:

From the information given that, a marketing firm asks participants to rate the effectiveness of ads that varied by length (short, long). That is, $p=2$, and by type of technology of (static, interactive). That is, $q=2$. Also, the sum of squares for length is 15, the sum of squares for length $\times$ technology is 144, the sum of squares for error is 570, the total sum of squares is 849, and the degrees of freedom for error are 114.

The formulas for computing the F table are,

Source of Variation	SS	df	MS	$F_{\text{obt}}$
Factor A	$S{S}_{\text{A}}$	$p-1$	$\frac{S{S}_{\text{A}}}{d{f}_{\text{A}}}$	$\frac{M{S}_{\text{A}}}{M{S}_{\text{E}}}$
Factor B	$S{S}_{\text{B}}$	$q-1$	$\frac{S{S}_{\text{B}}}{d{f}_{\text{B}}}$	$\frac{M{S}_{\text{B}}}{M{S}_{\text{E}}}$
$\text{A}\times \text{B}$	$S{S}_{\text{A}\times \text{B}}$	$\left(p-1\right)\left(q-1\right)$	$\frac{S{S}_{\text{A}\times \text{B}}}{d{f}_{\text{A}\times \text{B}}}$	$\frac{M{S}_{\text{A}\times \text{B}}}{M{S}_{\text{E}}}$
Error (within groups)	$S{S}_{\text{E}}$	$pq\left(n-1\right)$	$\frac{S{S}_{\text{E}}}{d{f}_{\text{E}}}$
Total	$S{S}_{\text{T}}$	$npq-1$

Substituting $p=2$ in the degrees of freedom for main effect length formula

$\begin{array}{c}d{f}_{\text{L}}=\left(2-1\right)\\ =1\end{array}$

The degrees of freedom for main effect length are 1.

Substituting $d{f}_{\text{L}}=1,S{S}_{\text{L}}=15$ in the mean square for main effect length formula

$\begin{array}{c}M{S}_{\text{L}}=\frac{15}{1}\\ =15\end{array}$

The mean square for main effect length is 15.

Substituting $q=2$ in the degrees of freedom for main effect technology formula

$\begin{array}{c}d{f}_{\text{TC}}=\left(2-1\right)\\ =1\end{array}$

The degrees of freedom for main effect technology are 1.

Substituting $S{S}_{\text{L}}=15,S{S}_{\text{L}\times \text{TC}}=144,S{S}_{\text{E}}=570,S{S}_{\text{T}}=849$ in the sum of squared for main effect technology formula

$\begin{array}{c}S{S}_{\text{TC}}=S{S}_{\text{T}}-\left(S{S}_{\text{L}}+S{S}_{\text{L}\times \text{TC}}+S{S}_{\text{E}}\right)\\ =849-\left(570+144+15\right)\\ =849-729\\ =120\end{array}$

The sum of squared for main effect technology is 120.

Substituting $d{f}_{\text{TC}}=1,S{S}_{\text{TC}}=120$ in the mean square for main effect technology formula

$\begin{array}{c}M{S}_{\text{TC}}=\frac{120}{1}\\ =120\end{array}$

The mean square for main effect technology is 120.

Substituting $p=2,q=2$ in the degrees of freedom for interaction length $\times$ technology formula

$\begin{array}{c}d{f}_{\text{L}\times \text{TC}}=\left(2-1\right)\left(2-1\right)\\ =1\times 1\\ =1\end{array}$

The degrees of freedom for interaction length $\times$ technology are 1.

Substituting $d{f}_{\text{L}\times \text{TC}}=1,S{S}_{\text{L}\times \text{TC}}=144$ in the mean square for interaction length $\times$ technology formula

$\begin{array}{c}M{S}_{\text{L}\times \text{TC}}=\frac{144}{1}\\ =144\end{array}$

The mean square for interaction length $\times$ technology is 144.

Substituting $d{f}_{\text{E}}=114,S{S}_{\text{E}}=570$ in the mean square for error formula

$\begin{array}{c}M{S}_{\text{E}}=\frac{570}{114}\\ =5\end{array}$

The mean square for error is 5.

Substituting $d{f}_{\text{L}}=1,d{f}_{\text{TC}}=2,d{f}_{\text{L}\times \text{TC}}=1,d{f}_{\text{E}}=114$ in the total degrees of freedom formula

$\begin{array}{c}d{f}_T=d{f}_{\text{L}}+d{f}_{\text{TC}}+d{f}_{\text{L}\times \text{TC}}+d{f}_{\text{E}}\\ =1+1+1+114\\ =117\end{array}$

The total degrees of freedom are 117.

Substituting $M{S}_{\text{L}}=15,M{S}_{\text{E}}=5$ in the F obtained value for length formula

$\begin{array}{c}{F}_{\text{L}}=\frac{15}{5}\\ =3\end{array}$

The F obtained value for length is 3.

Substituting $M{S}_{\text{TC}}=120,M{S}_{\text{E}}=5$ in the F obtained value for technology formula

$\begin{array}{c}{F}_{\text{TC}}=\frac{120}{5}\\ =24\end{array}$

The F obtained value for technology is 24.

Substituting $M{S}_{\text{L}\times \text{TC}}=144,M{S}_{\text{E}}=5$ in the F obtained value for interaction formula

$\begin{array}{c}{F}_{\text{L}\times \text{TC}}=\frac{144}{5}\\ =28.8\end{array}$

The F obtained value for interaction is 28.8.

The completed F table is,

Source of Variation	SS	df	MS	$F_{\text{obt}}$
Length	15	1	15	3.00
Technology	120	1	120	24.00
Length $\times$ Technology	144	1	144	28.80
Error	570	114	5
Total	849	117

Decision rules:

• If the test statistic value is greater than the critical value, then reject the null hypothesis
• If the test statistic value is smaller than the critical value, then retain the null hypothesis

Hypothesis test for main effect factor A (length):

Let ${\sigma }_{\mu \text{'}s}^2$ be the group means of the main effect length.

Null hypothesis:

$H_0:{\sigma }_{\mu \text{'}s}^2=0$

That is, the group means for main effect length do not significantly vary in the population.

Alternative hypothesis:

$H_0:{\sigma }_{\mu \text{'}s}^2>0$

That is, the group means for main effect length do significantly vary in the population.

Critical value:

The considered significance level is $\alpha =0.05$.

The degrees of freedom for numerator are 1, the degrees of freedom for denominator are 114 from completed F table.

From the Appendix C: Table C.3-Critical values for F distribution:

• Locate the value 1 in degrees of freedom numerator row.
• Locate the value 114 in degrees of freedom denominator row.This value is not in the table, consider the next highest value that is 120.
• Locate the 0.05 level of significance (value in lightface type) in combined row.
• The intersecting value that corresponds to the (1, 114) with level of significance 0.05 is 3.92.

Thus, the critical value for $df=\left(1,114\right)$ with 0.05, level of significance is 3.92.

Conclusion:

The value of test statistic is 3.00.

The critical value is 3.92.

The test statistic value is less than the critical value.

The test statistic value does not falls under critical region.

Hence the null hypothesis is retained.

The decision is the group means for main effect lengthdo significantly vary in the population.

Hypothesis test for main effect factor B (technology):

Let ${\sigma }_{\mu \text{'}s}^2$ be the group means of the main effect technology.

Null hypothesis:

$H_0:{\sigma }_{\mu \text{'}s}^2=0$

That is, the group means for main effect technology do not significantly vary in the population.

Alternative hypothesis:

$H_0:{\sigma }_{\mu \text{'}s}^2>0$

That is, the group means for main effect technology do significantly vary in the population.

Critical value:

The considered significance level is $\alpha =0.05$. The degrees of freedom for numerator are 1, the degrees of freedom for denominator are 114 from completed F table. From the Appendix C: Table C.3-Critical values for F distribution the critical value is 3.92.

Thus, the critical value for $df=\left(1,114\right)$ with 0.05, level of significance is 3.92.

Conclusion:

The value of test statistic is 24.00.

The critical value is 3.92.

The test statistic value is greater than the critical value.

The test statistic value falls under critical region.

Hence the null hypothesis is rejected.

The decision is the group means for main effect technology do significantly vary in the population.

Hypothesis test for interaction effect of factor A and B:

Let ${\sigma }_{\mu \text{'}s}^2$ be the group means of the interaction effect.

Null hypothesis:

$H_0:{\sigma }_{\mu \text{'}s}^2=0$

That is, the means of length do not significantly vary by technology or the combinations of these factors.

Alternative hypothesis:

$H_0:{\sigma }_{\mu \text{'}s}^2>0$

That is, the means of length do significantly vary by technology or the combinations of these factors.

Critical value:

The considered significance level is $\alpha =0.05$. The degrees of freedom for numerator are 1, the degrees of freedom for denominator are 114 from completed F table. From the Appendix C: Table C.3-Critical values for F distribution the critical value is 3.92.

Thus, the critical value for $df=\left(1,114\right)$ with 0.05, level of significance is 3.92.

Conclusion:

The value of test statistic is 28.8.

The critical value is 3.92.

The test statistic value is greater than the critical value.

The test statistic value falls under critical region.

Hence the null hypothesis is rejected.

The decision is the means of length do significantly vary by technology or the combinations of these factors.

2.

To determine

Explain the next step based on the results obtained in the test.

Answer to Problem 25CAP

The next step based on the results obtained in the test is that simple main effect tests for the significant interaction have to be calculated.

Explanation of Solution

Justification: The result of the test is that the main effect ‘technology’ is significant but another main effect ‘length’ is not significant. The interaction effect of length and technology is significant. If the interaction effect is significant in the study the next is to analyze the interaction using the simple main effect tests.

The test that is used for determining the interaction between the two factors is significant or not by comparing the mean differences or comparing the single main effect of one factor with each and every level of the second factor is termed as significant main effect test.