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	30	23
24	31	29
36	33	27
40	22	31
25	29	35

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.

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: μ_50°C ≠ μ_60°C ≠ μ_70°C    H₀: 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°C

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

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

p-value =

State your conclusion.

Reject H₀. There is 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 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.

Answer 1

(a)

Analysis of Variance Table is constructed as follows:

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F	p - value
Treatments	30	2	30/2=15.00	15.00/29.33 = 0.51	0.6122
Error	352	12	352/12=29.33
Total	382	14

Explanation:
The ntries in the Analysis of Variance Table are got as follows:

From the given data, the following Table is calculated:

	500 C	600 C	700 C
n	5	5	5
Sum	160	145	145
Mean	32	29	29
	5322	4375	4285
Std. Dev.	7.100	.4.183	4.472
SS	202	70	80

F = 15.00/29.33 = 0.51

By Technology, p - value = 0.6122

(b)

Correct option:

H_a: μ_50°C ≠ μ_60°C ≠ μ_70°C    H₀: μ_50°C = μ_60°C = μ_70°C

(c)

the value of the test statistic. is given by:

F = 15.00/29.33 = 0.51

(d)

p - value = 0.6122

(e)

Correct option:

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