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

August 30, 2017

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
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.
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.