Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Is honey a cough remedy?Refer to the Archives of Pediatrics and Adolescent Medicine(December 2007) study of honeyas a remedy for coughing, Exercise 2.31 (p. 86). The coughingimprovement scores (as determined by the children’s parents)for the patients in the over-the-counter cough medicinedosage (DM) group, honey dosage group, and control groupare reproduced in the accompanying table.

Honey Dosage

12, 11, 15, 11, 10, 13, 10, 4 ,15, 16, 9, 14, 10, 6, 10, 8, 11, 12, 12, 8, 12, 9, 11, 15, 10, 15, 9, 13, 8, 12, 10, 8, 9, 5, 12

DM Dosage

4, 6, 9, 4, 7, 7, 7, 9, 12 ,10, 11, 6, 3, 4, 9, 12, 7, 6, 8, 12, 12, 4, 12, 13, 7, 10

No Dosage
(Control)

13, 9, 4, 4, 10, 15, 9, 5, 8, 6, 1, 0, 8, 12, 8, 7, 7, 1, 6, 7, 7, 12, 7, 9, 7, 9, 5, 11, 9, 5, 6, 8, 8, 6, 7, 10, 9, 4, 8, 7, 3, 1, 4, 3

a.Find the standard deviation of the improvement scores for the honey dosage group.

b.Find the standard deviation of the improvement scores for the DM dosage group.

c.Find the standard deviation of the improvement scores for the control group.

d.Based on the results, parts a–c, which group appears to have the most variability in coughing improvement scores? The least variability?

Short Answer

Expert verified

a. Standard Deviation = 2.85

b. Standard Deviation = 3.037

c. Standard Deviation = 2.976

d. Most Variable = DM dosage group

Least Variable = Honey dosage group

Step by step solution

01

Finding the standard deviation for honey dosage group

Mean=SumofallscoresNo.ofscores=12+11+15+11+10+13+10+4+15+16+9+14+10+6+10+8+11+12+12+8+12+9+11+15+10+15+9+13+8+12+10+8+9+5+1235=37535=10.71

x

(x-x¯)

(x-x¯)2

12

1.29

1.66

11

0.29

0.08

15

4.29

18.40

11

0.29

0.08

10

-0.71

0.50

13

2.29

5.24

10

-0.71

0.50

4

-6.71

45.02

15

4.29

18.40

16

5.29

27.98

9

-1.71

2.92

14

3.29

10.82

10

-0.71

0.50

6

-4.71

22.18

10

-0.71

0.50

8

-2.71

7.34

11

0.29

0.08

12

1.29

1.66

12

1.29

1.66

8

-2.71

7.34

12

1.29

1.66

9

-1.71

2.92

11

0.29

0.08

15

4.29

18.40

10

-0.71

0.50

15

4.29

18.40

9

-1.71

2.92

13

2.29

5.24

8

-2.71

7.34

12

1.29

1.66

10

-0.71

0.50

8

-2.71

7.34

9

-1.71

2.92

5

-5.71

32.60

12

1.29

1.66

Sum

0

277.14

Variance=(x-x¯)2n-1=277.1435-1=277.1434=8.15StandardDeviation=Variance=8.15=2.85

Therefore, Standard deviation for honey dosage group is 2.85.

02

Calculating the standard deviation for DM dosage group

Mean=SumofallscoresNo.ofscores=4+6+9+4+7+7+7+9+12+10+11+6+3+4+9+12+7+6+8+12+12+4+12+13+7+1026=21126=8.115

x

(x-x¯)

(x-x¯)2

4

-4.115

16.93

6

-2.115

4.47

9

0.885

0.78

4

-4.115

16.93

7

-1.115

1.24

7

-1.115

1.24

7

-1.115

1.24

9

0.885

0.78

12

3.885

15.09

10

1.885

3.55

11

2.885

8.32

6

-2.115

4.47

3

-5.115

26.16

4

-4.115

16.93

9

0.885

0.78

12

3.885

15.09

7

-1.115

1.24

6

-2.115

4.47

8

-0.115

0.01

12

3.885

15.09

12

3.885

15.09

4

-4.115

16.93

12

3.885

15.09

13

4.885

23.86

7

-1.115

1.24

10

1.885

3.55

Sum

0

230.65

Variance=(x-x¯)2n-1=230.6526-1=230.6525=9.226StandardDeviation=Variance=9.226=3.037

Therefore, Standard deviation for DM dosage group is 3.03.

03

Finding the standard deviation for honey dosage group

Mean=SumofallscoresNo.ofscores=5+8+6+1+0+8+12+8+7+7+1+6+7+7+12+7+9+5+6+8+8+6+7+10+9+4+8+7+3+1+4+8+7+3+1+4+337=22337=6.027

x

(x-x¯)

(x-x¯)2

5

-1.027

1.05

8

1.973

3.89

6

-0.027

0.0007

1

-5.027

25.27

0

-6.027

36.32

8

1.973

3.89

12

5.973

35.67

8

1.973

3.89

7

0.973

0.94

7

0.973

0.94

1

-5.027

25.27

6

-0.027

0.0007

7

0.973

0.94

7

0.973

0.94

12

5.973

35.67

7

0.973

0.94

9

2.973

8.83

5

-1.027

1.05

6

-0.027

0.0007

8

1.973

3.89

8

1.973

3.89

6

-0.027

0.0007

7

0.973

0.94

10

3.973

15.78

9

2.973

8.83

4

-2.027

4.11

8

1.973

3.89

7

0.973

0.94

3

-3.027

9.16

1

-5.027

25.27

4

-2.027

4.11

8

1.973

3.89

7

0.973

0.94

3

-3.027

9.16

1

-5.027

25.27

4

-2.027

4.11

3

-3.027

9.16

Sum

0

318.973

Variance=(x-x¯)2n-1=318.97337-1=318.97336=8.86StandardDeviation=Variance=8.86=2.976

Therefore, Standard deviation for control group is 2.976.

04

Identifying the least and the most variable group

The most variable group is the DM dosage group as it has the highest standard deviation out of the 3.

The least variable group is the honey dosage group with the lowest standard deviation value.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the following three measurements: 0, 4, and 12. Find the z-score for each measurement if they are from a population with the following mean and standard deviation equal to

a.µ = 2 and σ = 1

b.µ = 4 and σ = 2

c.µ = 8 and σ = 2

d.µ = 8 and σ = 8

For each of the following data sets, compute xbar, s2, and s. If appropriate, specify the units in which your answers are expressed.

a.4, 6, 6, 5, 6, 7

b.- \(1, \)4, - \(3, \)0, - \(3, - \)6

c.3/5 %, 4/5 %, 2/5 %, 1/5 %, 1/16 %

d.Calculate the range of each data set in parts a–c.

Salaries of bachelor’s degree graduates. PayScale, Inc., an online provider of global compensation data, conducts an annual salary survey of bachelor’s degree graduates. Three of the many variables measured by PayScale are the graduate’s current salary, mid-career salary, and the college or university where they obtained their degree. Descriptive statistics are provided for each of the over 400 colleges and universities that graduates attended. For example, graduates of the University of South Florida (USF) had a mean current salary of \(57,000, a median mid-career salary of \)48,000, and a mid-career 90th percentile salary of $131,000. Describe the salary distribution of USF bachelor’s degree graduates by interpreting each of these summary statistics.

Hotels’ use of ecolabels.Ecolabels such as Energy Star, Green Key, and Audubon Internationalare used by hotels to advertise their energy-saving and conservation policies. The Journal of Vacation Marketing(January 2016) published a study to investigate how familiar travelers are with these ecolabels and whether travelers believe they are credible. A sample of 392 adult travelers were administered a questionnaire. One question showed a list of 6 different eco-labels, and asked, “How familiar are you with this ecolabel, on a scale of 1 (not familiar at all) to 5 (very familiar).” Summarized results for the numerical responses are given in the table.

a.Give a practical interpretation of the mean response for Energy Star.

b.Give a practical interpretation of the median response for Energy Star.

c.Give a practical interpretation of the response mode for Energy Star.

d.Based on these summary statistics, which ecolabel appears to be most familiar to travelers?

Ecolabel

Mean

Median

Mode

Energy Star

4.44

5

5

TripAdvisor

3.57

4

4

Green Leaders Audubon

2.41

2

1

International U.S Green

2.28

2

1

Building Council Green Business

2.25

2

1

Green Key

2.01

1

1

Rankings of research universities.Refer to the College Choice 2015 Rankings of National Research Universities, Exercise 2.43 (p. 95). Recall that data on academic reputation score, financial aid awarded, and net cost to attend for the top 50 research universities are saved in the TOPUNIV file. The 50 academic reputation scores are listed in the accompanying table.

99 92 94 95 97 91 91 92 92 89 84 85 100 87 83 83 89 79 94 79 79 87 76 67 76 76 76 70 74 64 74 69 66 72 65 76 64 65 61 69 62 69 52 64 64 47 60 57 63 62

a.Find the median, lower quartile, and upper quartile for the data.

b.Find IQR for the data.

c.Graph the data with a box plot.

d.Do you detect any outliers? Suspect outliers?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free