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	31	22
25	31	29
35	33	29
39	23	29
26	27	36

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

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.)

t stat =

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

p-value =

State your conclusion.

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.

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

Answer 1

one way anova-

treatment	50°C	60°C	70°C
count, ni =	5	5	5
mean , x̅ i =	32.000	29.00	29.000
std. dev., si =	6.164	4.000	4.950
sample variances, si^2 =	38.000	16.000	24.500
total sum	160	145	145		450	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =				30.00
square of deviation of sample mean from grand mean,( x̅ - x̅̅)²	4.000	1.000	1.000
					TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	20.000	5.000	5.000		30
SS(within ) = SSW = Σ(n-1)s² =	152.000	64.000	98.000		314.000

no. of treatment , k =   3
df between = k-1 =    2
N = Σn =   15
df within = N-k =   12
  
mean square between groups , MSB = SSB/k-1 =    15.0000
  
mean square within groups , MSW = SSW/N-k =    26.1667
  
F-stat = MSB/MSW =    0.5732

anova table
	SS	df	MS	F	p-value	F-critical
treatments	30.00	2	15.00	0.57	0.5784	3.885
error	314.00	12	26.17
Total:	344.00	14
					α =	0.05

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

test stat = 0.57

p value=0.5784

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