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 30 26 26 20 27 31 32 30 31 35 19 33 28 23 34 a. Construct an analysis of variance table (to 2 decimals but p-value to 4 decimals, if necessary). Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Treatments Error Total b. Use a level of significance to test whether the temperature level has an effect on the mean yield of the process. Calculate the value of the test statistic (to 2 decimals). The -value is - Select your answer -less than .01between .01 and .025between .025 and .05between .05 and .10greater than .10 What is your conclusion? - Select your answer -Conclude that the mean yields for the three temperatures are not all equal Do not reject the assumption that the mean yields for the three temperatures are equal

Answer 1

Solution:

a) The null and alternative hypothesis are as follows:

i.e. The mean yields for the three temperatures are equal.

H_{​​​​​​1} : The mean yields for the three temperatures are not all equal.

To test the hypothesis we shall perform one way Anova. The test statistic is given as follows:

Where, and

If we have K treatments and total number of observations is N then,

df_{​​​​​treatment} = K - 1 and df_{​​​​​error} = N - K

We calculate different terms in one way Anova as follows:

Raw sum of square :

Correction Factor :

Where, and N is total number of observations.

N = 15

Total Sum of square :

TSS = RSS - CF

TSS = 12371 - 12041.6667 = 329.3333

Sum of squares due to treatment :

Where, T_{​​​​​​1.} , T_{​​​​​​2.} and T_{​​​​​​3.} are column sum for each of the temperature groups. n1, n2 and n3 are the number of observations in each groups.

Sum of square due to error :

Since, we have 3 treatment groups and N = 15, therefore

df_{​​​​​treatment} = (3 - 1) = 2 and df_{​​​​​error} = (15 - 3) = 12

Mean square due to treatment :

Mean square due to error :

F test statistic :

The p-value for the test statistic is given as follows:

p-value = P(F > value of the test statistic)

p-value = P(F > 2.37288)

p-value = 0.1354

The Anova table is given below:

Source of variation	Sum of square	Degrees of freedom	Mean square	F	p-value
Treatment	93.33	2	46.67	2.37	0.1354
Error	236	12	19.67
Total	329.33	14

b)

We shall use significance level of 0.10.

We make decision rule as follows:

If p-value is greater than the significance level, the we fail to reject H_{​​​​​​0} at given significance level.

If p-value is less than the significance level, the we reject H_{​​​​​​0} at given significance level.

Since, p-value is greater than the significance level of 0.10, therefore we shall be fail to reject the null hypothesis (H_{​​​​​​0}).

Conclusion: There is not sufficient evidence to reject the assumption that the mean yields for the three temperatures are equal.

Or in other words, Do not reject the assumption that the mean yields for the three temperatures are equal.

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