The summary from the data.

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Answer 1

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Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 12, Problem 64E

(a)

To determine

To find: The summary from the data.

(a)

Expert Solution

Answer to Problem 64E

Solution: The summary from the data is as shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 1

Explanation of Solution

Calculation: To draw the inference from the data, use Excel. Steps are provided below:

Step 1: Open Excel sheet and write the value for three treatments. The screenshot is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 2

Step 2: The sample size for “Novels” can be obtained by using the formula =COUNT(A2:A68). The screenshot is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 3

The sample size for “Poems” can be obtained by using the formula =COUNT(B2:B33). The screenshot is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 4

The sample size for “Nonfiction” can be obtained by using the formula =COUNT(C2:C25). The screenshot is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 5

Step 3: The average value for “Novels” can be obtained by using the formula =AVERAGE(A2:A68). The screenshot is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 6

The average value for “Poems” can be obtained by using the formula =AVERAGE(B2:B33). The screenshot is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 7

The average value for “Nonfiction” can be obtained by using the formula =AVERAGE(C2:C25). The screenshot is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 8

Step 4: The standard deviation for “Novels” can be obtained by using the formula =STDEV(A2:A68). The screenshot is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 9

The standard deviation for “Poems” can be obtained by using the formula =STDEV(B2:B33). The screenshot is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 10

The standard deviation for “Nonfiction” can be obtained by using the formula =STDEV(C2:C25). The screenshot is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 11

Step 5: The standard error for “Novels” can be obtained by using the formula =STDEV(A2:A68)/(SQRT(COUNT(A2:A68))). The screenshot is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 12

The standard error for “Poems” can be obtained by using the formula =STDEV(B2:B33)/(SQRT(COUNT(B2:B33))). The screenshot is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 13

The standard error for “Nonfiction” can be obtained by using the formula =STDEV(C2:C25)/(SQRT(COUNT(C2:C25))). The screenshot is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 14

The result is obtained.

The screenshot of the summary of results is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 15

(b)

To determine

Necessary assumption to carry out ANOVA test.

(b)

Expert Solution

Answer to Problem 64E

Solution: All the distribution are approximately normal and pooling of standard deviation is applicable.

Explanation of Solution

The necessary assumption is that all the treatment values are approximately normally distributed about their means and the single or pooled standard deviation can be used for all the treatments such that σmaximum<2×σminimum.

(c)

To determine

To test: The ANOVA model.

(c)

Expert Solution

Answer to Problem 64E

Solution: The required ANOVA model is obtained as follows:

Degree of freedom=(2,117)P−value=0.00088F=7.476

Here, the F value is greater than the F critical value. Hence, the null hypothesis can be rejected significantly.

Explanation of Solution

Calculation: Single factor ANOVA model is performed by following these steps:

Step 1: Open Excel sheet and write the value for three treatments. The screenshot is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 16

Step 3: Data > Data Analysis > OK. The screenshot is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 17

Step 4: Select ANOVA: Single Factor > OK. The screenshot is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 18

Step 5: Select input range > OK. The screenshot is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 19

The model is obtained. The screenshot is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 20

Conclusion: Here, the F value is greater than the F critical. Hence, null hypothesis can be rejected significantly.

(d)

To determine

To test: The contrast to compare the poets with the writers.

(d)

Expert Solution

Answer to Problem 64E

Solution: The P value is P(t<−3.8)<.0001, which is much less than 0 .05 significance level, hence null hypothesis can be rejected at 5% significance level.

Explanation of Solution

A contrast (c) is defined as a combination of population means.

c=∑i=1nai×μi

where ∑i=1nai=0 and μi is the sample means.

The null hypothesis, H0:c=0, can be tested by calculating the t statistics as follows:

t=cSEc

where SEc=sp×(a1)2n1+(a2)2n2+(a3)2n3 the standard error of contrast and c is the contrast.

The null hypothesis is tested to compare novelists with the nonfiction writers. The following steps are followed with reference to part (c) on excel to test the null hypothesis:

Step 1: Contrast c is calculated as follows:

c=∑i=1nai×μi=(−0.5×71.6667)+(−0.5×77)+(1×62.3871)=−11.9462

Step 2: The pooled standard deviation is obtained as follows:

sp=(n1−1)s12+(n2−1)s22+(n3−1)s32n1+n2+n3−3=(66−1)169.703+(23−1)207.364+(31−1)287.97866+23+31−3=14.4766

Step 3: Standard error of contrast is obtained as follows:

SEc=sp×(−0.5)2n1+(−0.5)2n2+(1)2n3=13.3875×(−0.5)266+(−0.5)223+(1)231=3.13564

Step 4: The resultant t statistic can be calculated as follows:

t=cSEc=−11.94623.13564=−3.80983

(e)

To determine

To test: A contrast to compare the novelists with the nonfiction writers.

(e)

Expert Solution

Answer to Problem 64E

Solution: The p- value is P(t<−1.65)=.0505 that is approximately equal to 0.05 significance level, hence null hypothesis can be rejected at 5% significance level.

Explanation of Solution

The null hypothesis is tested to compare novelists with the nonfiction writers. The following steps are followed with reference to part (c) on excel to test the null hypothesis:

Step 1: Contrast c is calculated as follows:

c=∑i=1nai×μi=(−0.5×71.6667)+(0.5×77)=2.66667

Step 2: The pooled standard deviation is obtained as follows:

sp=(n1−1)s12+(n2−1)s22n1+n2−2=(66−1)169.703+(23−1)14.400166+23−2=13.3875

Step 3: Standard error of contrast can be calculated as follows:

SEc=sp×(−0.5)2n1+(−0.5)2n2=13.3875×(−0.5)266+(−0.5)223=1.6208

Step 4: The resultant t statistic can be calculated as follows:

t=cSEc=2.666671.6208=1.64528

Conclusion: The p- value is P(t<−1.65)=.0505 that is approximately equal to 0.05 significance level, hence null hypothesis can be rejected at 5% significance level.

(f)

To determine

To test: The multiple comparison t test.

(f)

Expert Solution

Answer to Problem 64E

Solution: The poem’s mean is significantly different from other means.

Explanation of Solution

To perform the multiple comparison t-test, there are three possible combinations that are novels versus poems, poems versus nonfiction, and novels versus nonfiction. The steps to be followed on excel to obtain the multiple comparison t test are mentioned below.

Calculation:

Multiple comparison t-test is performed by following these steps:

Step 1: Open Excel sheet and write the value for three treatments. The screenshot is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 21

Step 2: Data > Data Analysis > OK. The screenshot is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 22

Step 3: Select t-test: Two-Sample Assuming Equal Variances > OK. The screenshot is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 23

Step 4: Select input range > OK. The screenshot is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 24

Step 5: Repeat step 5 to compare low dose versus high dose and control versus high dose.

The result for comparison of novels versus poems is obtained as follows:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 25

The result for comparison of poems versus nonfiction is obtained as follows:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 26

The result for comparison of novels versus nonfiction is obtained as follows:

Introduction to the Practice of Statistics, Chapter 12, Problem 64E , additional homework tip 27

Conclusion: From the above comparison, the P value for two-tail for poems versus novels and poems versus nonfiction is less than 0.05 at 5% significance level. Hence, poem’s mean is significantly different from other means.

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