# Analysis of variance

Please see the attached questions.

Week 3 Revised E-text Assignments: (4 points)

Directions: You may include the statistical software output, but you must also include a well-written explanation of the findings. Be sure to answer the question asked in each problem, and explain why, with reference to your output. If you calculate the answers manually, be sure to show your work. I would prefer a Word document with your answers below each problem, but you may also submit an Excel document.

10.44: Does lovastatin (a cholesterol-lowering drug) reduce the risk of heart attack? In a Texas study, researchers gave lovastatin to 2,325 people and an inactive substitute to 2,081 people (average age 58). After 5 years, 57 of the lovastatin group had suffered a heart attack, compared with 97 for the inactive pill. (a) State the appropriate hypotheses. (b) Obtain a test statistic and p-value. Interpret the results at α = .01. (c) Is normality assured? (d) Is the difference large enough to be important? (e) What else would medical researchers need to know before prescribing this drug widely?

Hypothesis test for two independent proportions

p1 p2 pc

0.0245 0.0466 0.0349 p (as decimal)

57/2325 97/2081 154/4406 p (as fraction)

56.963 96.975 153.937 X

2325 2081 4406 n

-0.0221 difference

0. hypothesized difference

0.0055 std. error

-3.99 z

3.33E-05 p-value (one-tailed, lower)

-0.0366 confidence interval 99.% lower

-0.0076 confidence interval 99.% upper

0.0145 half-width

2: Lester Hollar is Vice President for Human Resources for a large manufacturing company. In recent years he has noticed an increase in absenteeism that he thinks is related to the general health of the employees. Four years ago, in an attempt to improve the situation, he began a fitness program in which employees exercise during their lunch hour. To evaluate the program, he selected a random sample of eight participants and found the number of days each was absent in the six months before the exercise program began and in the last six months. Below are the results. At the .05 significance level, can he conclude that the number of absences has declined? Explain.

Employee Before After

1 6 4

2 7 2

3 6 1

4 7 4

5 6 3

6 5 7

7 4 2

8 5 4

ANOVA

11.24 In a bumper test, three types of autos were deliberately crashed into a barrier at 5 mph, and the resulting damage (in dollars) was estimated. Five test vehicles of each type were crashed, with the results shown below. Research question: Are the mean crash damages the same or significantly different for these three vehicles? Are there any significant differences between groups? Explain.

Crash Damage in Dollars

Goliath Varmint Weasel

1100 1290 1020

760 1400 2100

980 1390 1830

975 1850 1250

750 1100 1920

One factor ANOVA

Mean n Std. Dev

913.0 5 152.71 Goliath

1,406.0 5 275.92 Varmint

1,624.0 5 464.04 Weasel

1,314.3 15 429.78 Total

ANOVA table

Source SS df MS F p-value

Treatment 1,326,823.33 2 663,411.667 6.32 .0133

Error 1,259,120.00 12 104,926.667

Total 2,585,943.33 14

Post hoc analysis

Tukey simultaneous comparison t-values (d.f. = 12)

Goliath Varmint Weasel

913.0 1,406.0 1,624.0

Goliath 913.0

Varmint 1,406.0 2.41

Weasel 1,624.0 3.47 1.06

critical values for experimentwise error rate:

0.05 2.67

0.01 3.56

p-values for pairwise t-tests

Goliath Varmint Weasel

913.0 1,406.0 1,624.0

Goliath 913.0

Varmint 1,406.0 .0331

Weasel 1,624.0 .0046 .3082

4. There are three hospitals in the Fayetteville, Arkansas area. The following data show the number of outpatient surgeries performed at each hospital last week. At the .05 significance level, can we conclude there is a difference in the mean number of surgeries performed by hospital or by day of the week?

Number of Surgeries Performed

St. Luke's St. Vincent's Mercy

Monday 14 18 24

Tuesday 20 24 14

Wednesday 16 22 14

Thursday 18 20 22

Friday 20 28 24

Randomized blocks ANOVA

Mean n Std. Dev

19.86666667 17.600 5 2.608 Treatment 1

19.86666667 22.400 5 3.847 Treatment 2

19.86666667 19.600 5 5.177 Treatment 3

18.667 3 5.033 Block 1

19.333 3 5.033 Block 2

17.333 3 4.163 Block 3

20.000 3 2.000 Block 4

24.000 3 4.000 Block 5

19.867 15 4.240 Total

ANOVA table

Source SS df MS F p-value

Treatments 58.13 2 29.067 1.97 .2011

Blocks 75.73 4 18.933 1.29 .3524

Error 117.87 8 14.733

Total 251.73 14

#### Solution Preview

Please see the attachments.

Week 3 Revised E-text Assignments: (4 points)

Directions: You may include the statistical software output, but you must also include a well-written explanation of the findings. Be sure to answer the question asked in each problem, and explain why, with reference to your output. If you calculate the answers manually, be sure to show your work. I would prefer a Word document with your answers below each problem, but you may also submit an Excel document.

10.44: Does lovastatin (a cholesterol-lowering drug) reduce the risk of heart attack? In a Texas study, researchers gave lovastatin to 2,325 people and an inactive substitute to 2,081 people (average age 58). After 5 years, 57 of the lovastatin group had suffered a heart attack, compared with 97 for the inactive pill. (a) State the appropriate hypotheses. (b) Obtain a test statistic and p-value. Interpret the results at α = .01. (c) Is normality assured? (d) Is the difference large enough to be important? (e) What else would medical researchers need to know before prescribing this drug widely?

Answer

The null hypothesis tested is

H0: There is no significant difference in proportion of heart attack in the two groups.

H1: The proportion of heart attacks in the lovastatin group is less than the inactive pill group.

The test Statistic used is

where

Rejection criteria: Reject the null hypothesis, if the observed significance ( p value ) is less than the significance level (0.01)

Details

Hypothesis test for two independent proportions

p1 p2 pc

0.0245 0.0466 0.0349 p (as decimal)

57/2325 97/2081 154/4406 p (as fraction)

56.963 96.975 153.937 X

2325 2081 4406 n

-0.0221 difference

0. hypothesized difference

0.0055 std. error

-3.99 z

3.33E-05 p-value (one-tailed, lower)

-0.0366 confidence interval 99.% lower

-0.0076 confidence interval 99.% upper

0.0145 half-width

Conclusion : Reject the null hypothesis. The sample provides enough evidence to support the claim ...

#### Solution Summary

Step by step method for computing test statistic for One way ANOVA is given in the answer. Interactive excel sheet is included. The user can edit the inputs and obtain the complete results for a new set of data.