Question

A chemist wishes to test the effect of four chemical agents on the strength of a particular type of cloth. Because there might be variability from one bolt to another, the chemist decides to use a randomized block design, with the bolts of cloth considered as blocks. She selects five bolts and applies all four chemicals in random order to each bolt. The resulting tensile strengths follow. Analyze the data from this experiment (use α = 0.05) and draw appropriate conclusions.

Bolt


Chemical 1 2 3 4 5


1 73 68 73 71 67


2 73 67 75 72 70


3 75 68 78 73 68


4 73 71 75 75 69

Answer 1

p > α

0.7038 > 0.05

Also since F < F critical

0.475 < 3.238

We failed to reject H₀

We do not have significant evidence at the given significance level to show that there is a difference among the four chemical agents on the strength of a particular type of cloth.

Explanation:

We are given that a chemist wishes to test the effect of four chemical agents on the strength of a particular type of cloth.

Since we are given data for four independent chemical agents to determine the effect on the strength of a particular type of cloth, therefore, a one-way analysis of variance may be used for the given problem.

ANOVA:

The one-way analysis of variance (ANOVA) may be used to find out whether there is any significant difference between the means of two or more independent categories of data.

Set up hypotheses:

Null hypotheses = H₀: μ₁ = μ₂ = μ₃ = μ₄

Alternate hypotheses = H₁: μ₁ ≠ μ₂ ≠ μ₃ ≠ μ₄

Set up decision rule:

We Reject H₀ if p ≤ α

OR

We Reject H₀ if F > F critical

ANOVA in Excel:

Step 1:

In the data tab, select data analysis

Step 2:

Select "Anova single factor" from the analysis tools

Step 3:

Select the destination of input data in the "input range"

Step 4:

Select "rows" for the option "Group By"

Step 5:

Tick the option "labels in first row"

Step 6:

Set alpha = 0.05

Step 7:

Select the destination of output data in the "output options"

Conclusion:

Please refer to the attached results.

The p-value is found to be

p = 0.7038

The F value is found to be

F = 0.475

The F critical value is found to be

F critical = 3.238

Since p > α

0.7038 > 0.05

We failed to reject H₀

Also since F < F critical

0.475 < 3.238

We failed to reject H₀

We do not have significant evidence at the given significance level to show that there is a difference among the four chemical agents on the strength of a particular type of cloth.