Question

To study the effect of temperature on yield in a chemical process, five batches were produced at each of three temperature levels. The results follow.

Temperature
50°C	60°C	70°C
35	29	22
24	32	27
35	34	27
40	23	31
26	27	38

1. Construct an analysis of variance table. (Round your values for MSE and F to two decimal places, and your p-value to four decimal places.)

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

Use a 0.05 level of significance to test whether the temperature level has an effect on the mean yield of the process.

2. State the null and alternative hypotheses.

H₀: μ_50°C = μ_60°C = μ_70°C
H_a: μ_50°C ≠ μ_60°C ≠ μ_70°CH₀: μ_50°C = μ_60°C = μ_70°C
H_a: Not all the population means are equal.    H₀: At least two of the population means are equal.
H_a: At least two of the population means are different.H₀: Not all the population means are equal.
H_a: μ_50°C = μ_60°C = μ_70°CH₀: μ_50°C ≠ μ_60°C ≠ μ_70°C
H_a: μ_50°C = μ_60°C = μ_70°C

3. Find the value of the test statistic. (Round your answer to two decimal places.)

4. Find the p-value. (Round your answer to four decimal places.)

p-value =

5. State your conclusion.

Do not reject H₀. There is not sufficient evidence to conclude that the mean yields for the three temperatures are not equal.

Reject H₀. There is not sufficient evidence to conclude that the mean yields for the three temperatures are not equal.    

Do not reject H₀. There is sufficient evidence to conclude that the mean yields for the three temperatures are not equal.

Reject H₀. There is sufficient evidence to conclude that the mean yields for the three temperatures are not equal.

Answer 1

The statistical software output for this problem is:

Hence,

1. ANOVA table:

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F	p-value
Treatments	30	2	15	0.45	0.6466
Error	398	12	33.17
Total	428	14

2. H₀: μ_50°C = μ_60°C = μ_70°C
H_a: Not all the population means are equal.

3. Test statistic = 0.45

4. P - value = 0.6466

5. Do not reject H₀. There is not sufficient evidence to conclude that the mean yields for the three temperatures are not equal.