# STATS Chapter 13: Problems 1, 5, 9, and 11

Question

The online problems correspond to these problems from the hard copy of the text:

Chapter 13: Problems 1, 5, 9, and 11

Notes:

For problem 1, in the data table you are given ????????2 (the sample variance) not ???????? like we have in the class

notes,

Also for problem 1, for part f on the online homework, leave the p-value selection blank and use the

critical value to make your conclusion. I’ll manually adjust the grades.

For problem 5, again leave the p-value selection blank and use the critical value to form your conclusion.

For problem 9, feel free to use Excel to analyze the data. Again, you don’t have to find the p-value for

the online homework.

For problem 11, you should use Excel to answer this question, and you should find the p-value. The

dataset, Paint.xlsx, is posted on K-State online. You may want to do the Excel Project before doing this

problem.

1. – PLEASE READ INSTRUCTIONS AFTER THIS QUESTION BEFORE ATTEMPTING TO ANSWER IT.

The following are notes provided for answering question 1 above. I

included them in the event it helps you solve the other problems.

We could see here that we have three treatments. Sample size is

six. The sample mean has been computed for each treatment, and the

sample variance.

Compute the sum of squares between treatments.

The sum of squares between treatments is given by the function the

sum as J goes from one to K, K being the number of treatments, and

for each of these treatments we want to multiply sample size times

the square deviation of the sample mean, minus the overall mean. To

compute the overall mean, we need to take each of the sample means

at 156, 142, 134 and divide by 3. Thus, the overall mean or X

double bar is 144. Continuing to substitute values in the equation,

the sum of square treatment is sample size of six, times 156, minus

144 squared for the first treatment group A, plus six times 142,

minus 144 squared for treatment B, plus six times 134 minus 144

squared for treatment C. Intermediate calculations reveals six

times 144, plus six times four, plus six times 100, which equals

1,488. The sum of square treatments is 1,488. The number of

treatments, K, is three.

Compute the mean square between treatments. The mean square is

obtained by dividing the sum of square between treatments by its

associated degrees of freedom or K minus 1. Thus, the mean square

treatment is 1,488 divided by two or 744.

Compute the sum of squares due to error. The sum of squares due to

error is given by the formula for each treatment, as J goes from

one to K. We want to multiply sample size minus 1, times the

treatment variance. Substituting, we obtain 5 times 154.4,

treatment A variance, plus 5 times 131.2, which is treatment B’s

variance, plus 5 times 110.4, treatment C’s variance. Continuing

the calculations, we get 822 plus 656, plus 552 or 2,030. The sum

of squared error is 2,030.

To compute the mean square due to error, we divide the sum of

square error by its associated degrees of freedom. In this case, N

sub-T, which is the total number of samples, 6 plus, plus 6, plus

6, or 18, minus 3, thus, the mean square error is 2,030 divided by

18 minus 3, which equals 135.3.

Set up the ANOVA table for this problem. In the ANOVA table, we

have the source of variation column where we look at the

treatments, the error and the total, the sum of squares, the

associated degrees of freedom, mean square, the F-value, the test

statistic and the associated P-value. To complete the table, when

the source of variation is treatments, the degrees of freedom is K

minus 1 or 3 minus 1 which is 2. The degrees of freedom for the

error is N sub-T minus K. The total number of the sample which is

18 minus K is 3 or 15. The total degrees of freedom is given by

total sample size minus 1 or 18 minus 1. Also, additively, we can

see that two plus 15 yields a total of 17. Computing the F

statistic, F is computed as the mean square treatment divided by

the mean square error or 744 over 135.3 which is equal to 5.50.

Updating the table, we see we also can compute the sum of square’s

total, as the sum of squares treatment plus some of square error.

Finally, it’s necessary to compute the P-value. You can see the Pvalue

posted here, which is obtained by using an Excel function,

the exact P-value, for demonstration purposes, we will show how to

go into the F table and compute the P-value. Degrees of freedom in

the numerator is 2, in the denominator, is 15. Looking down the

associated cells, we see that the value 5.50 is between .01 and

.025 which is validated by the exact P-value of .0162. From the F

table, the P-value is between .01 and .025.

At the alpha equal .05 level, test whether the means for the three

treatments are equal. If the P-value is less than the alpha of .05,

we can reject the null hypothesis that the means for the three

treatments are equal.

Because the P-value is less than alpha equal .05 or exactly .0162,

we would reject the hypothesis that the means of the three

treatments are equal.

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