Question

{Exercise 13.07}

Three different methods for assembling a product were proposed by an industrial engineer. To investigate the number of units assembled correctly with each method, 30 employees were randomly selected and randomly assigned to the three proposed methods in such a way that each method was used by 10 workers. The number of units assembled correctly was recorded, and the analysis of variance procedure was applied to the resulting data set. The following results were obtained: SST = 10,800; SSTR = 4560.

1. Set up the ANOVA table for this problem (to 2 decimals, if necessary).
   
   

   Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F
   Treatments
   Error
   Total

   
   
   
2. Use = .05 to test for any significant difference in the means for the three assembly methods.
   
   Calculate the value of the test statistic (to 2 decimals).
   
   
   The p-value is Selectless than .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 11
   
   What is your conclusion?
   SelectConclude not all means of the three assembly methods are equalCannot reject the assumption that the means of all three assembly methods are equal

Item 12are my calculations correct?

Thanks!

Answer 1

Concepts and reason

One-factor ANOVA model: In one factor ANOVA model only one factor is studied, that is only the effect on one factor is studied simultaneously. In this model there would be one dependent variable and one independent variable (with more than 2 levels). The analysis of variance test is used to compare the means of more than two groups.

Assumptions for ANOVA:

• Randomness and Independence: Samples are taken at random and are independent of each other.

• Normality: Samples are taken from a normally distributed population.

• Homogeneity of variance: The variances of probability distributions are equal.

Null hypothesis: The null hypothesis states that there is no difference in the test, which is denoted by ${H_0}$ . Moreover, the sign of null hypothesis is equal $\left( = \right)$ , greater than or equal $\left( \ge \right)$ and less than or equal $\left( \le \right)$ .

Alternative hypothesis: The hypothesis that differs from the ${H_\alpha }$ is called alternative hypothesis. This signifies that there is a significant difference in the test. The sign of alternative hypothesis is less than $\left( < \right)$ , greater than $\left( > \right)$ , or not equal $\left( \ne \right)$ .

Fundamentals

The One-Way ANOVA table is shown below:

The formula for SSE is, $SSE = SST - SSTR$ .

The formula for degrees of freedom:

$d{f_{{\rm{SSTR}}}} = r - 1$ , where r is the different methods for treatment

$d{f_{\rm{T}}} = n - 1$ , where n is the total number of observations

$d{f_{{\rm{SSE}}}} = n - r$

The formula for mean sum of squares:

$\begin{array}{l}\\MSTR = \frac{{SSTR}}{{d{f_{{\rm{SSTR}}}}}}\\\\MSE = \frac{{SSE}}{{d{f_{{\rm{SSE}}}}}}\\\end{array}$

The F ratio is, ${F_{TR}} = \frac{{MSTR}}{{MSE}}$

Rejection rule for p-value:

If $p{\rm{ - value}} \le \alpha \left( { = 0.05} \right)$ then reject the null hypothesis.

(1)

The sum of squares for error is obtained as shown below:

From information given, $SST = 10,800$ and $SSTR = 4,560$ .

$\begin{array}{c}\\SSE = SST - SSTR\\\\ = 10,800 - 4,560\\\\ = 6,240\\\end{array}$

From information given, there are three different methods for assembling a product, that is $r = 3$ . Also, 30 employees are randomly selected, that is $n = 30$ .

The degrees of freedom for treatments is,

$\begin{array}{c}\\d{f_{{\rm{SSTR}}}} = r - 1\\\\ = 3 - 1\\\\ = 2\\\end{array}$

The degrees of freedom for error is,

$\begin{array}{c}\\d{f_{{\rm{SSE}}}} = n - r\\\\ = 30 - 3\\\\ = 27\\\end{array}$

The degrees of freedom for total is,

$\begin{array}{c}\\d{f_{\rm{T}}} = n - 1\\\\ = 30 - 1\\\\ = 29\\\end{array}$

The mean sum of square for treatments is,

$\begin{array}{c}\\MSTR = \frac{{SSTR}}{{d{f_{{\rm{SSTR}}}}}}\\\\ = \frac{{4,560}}{2}\\\\ = 2,280\\\end{array}$

The mean sum of square for error is,

$\begin{array}{c}\\MSE = \frac{{SSE}}{{d{f_{{\rm{SSE}}}}}}\\\\ = \frac{{6,240}}{{27}}\\\\ = 231.11\\\end{array}$

The F ratio is obtained as shown below:

$\begin{array}{c}\\{F_{TR}} = \frac{{MSTR}}{{MSE}}\\\\ = \frac{{2,280}}{{231.11}}\\\\ = 9.87\\\end{array}$

The ANOVA table is,

(2.1)

State the hypotheses.

Let ${\mu _1}$ denote the mean for first method, ${\mu _2}$ denote the mean for second method and ${\mu _3}$ denote the mean for third method.

Null hypothesis:

${H_0}:{\mu _1} = {\mu _2} = {\mu _3}$

Alternative hypothesis:

${H_0}:{\mu _1} \ne {\mu _2} \ne {\mu _3}$

The F ratio is obtained as shown below:

$\begin{array}{c}\\{F_{TR}} = \frac{{MSTR}}{{MSE}}\\\\ = \frac{{2,280}}{{231.11}}\\\\ = 9.87\\\end{array}$

(2.2)

The p-value is obtained as shown below:

The test statistic value is 9.87, the numerator degrees of freedom is 2 and the denominator degrees of freedom is 27.

In ‘F distribution table’ locate the column with ‘2’ under ‘numerator degrees of freedom’. Locate the row with number ‘27’ under ‘denominator degrees of freedom’.

Locate the value that is close to the test statistic value. The value closer to test statistic $\left( { = 9.87} \right)$ value is 9.02. The p-value corresponding to the value 9.02 is 0.001.

Thus, the p-value (= 0.001) is less than 0.01.

(2.3)

The conclusion is stated below:

Use the level of significance, $\alpha = 0.05$ .

The p-value is less than 0.05.

Here, the p-value is less than level of significance.

That is, $p{\rm{ - value}}\left( { = {\rm{less}}\,{\rm{than}}\,0.01} \right) < \alpha \left( { = 0.05} \right)$

By, the rejection rule, the null hypothesis is rejected.

Therefore, it can be concluded that, there is evidence that the mean for three methods is significantly different.

Ans: Part 1

The ANOVA table is,

Part 2.1

The value of the test statistic is 9.87.

Part 2.2

The p-value is less than 0.01.

Part 2.3

Reject the null hypothesis, there is evidence that the mean for three methods is significantly different.