STAT 200 Week 6 Homework Problems

9.1.2

Many high school students take the AP tests in different subject areas.  In 2007, of the 144,796 students who took the biology exam, 84,199 of them were female.  In that same year, of the 211,693 students who took the Calculus AB exam 102,598 of them were female ("AP exam scores," 2013).  Estimate the difference in the proportion of female students taking the biology exam and female students taking the calculus AB exam using a 90% confidence level.

Suppose that, P1 = proportion of female students in biology exam

P2 = proportion of female students in Calculus exam.

Now the sample proportion for both samples as:

p1 = 84199/144796 = 0.5815

p2 = 102598/211693 = 0.4847

After that we need to calculate the pooled proportion p = (x1 + x2)/(n1 + n2)

Pooled proportion p = (84199 + 102598)/(144796 + 211693) = 0.524

Critical z for 90% confidence level = 1.645

E = critical z * standard error

Margin of error E = 1.645 * 0.524(1-0.524)(1144796+1211693)=0.0028

Hence 90% confidence interval

= ((0.5815 – 0.4847) – 0.0028, (0.5815 – 0.4847) + 0.0028)

= (0.094, 0.0996)

Hence, we can be 90% confident that the true difference in the proportion of female students taking the biology exam and female students taking the calculus AB exam will lie within (0.094, 0.0996)

 

9.1.5

Are there more children diagnosed with Autism Spectrum Disorder (ASD) in states that have larger urban areas over states that are mostly rural?  In the state of Pennsylvania, a fairly urban state, there are 245 eight-year-old diagnosed with ASD out of 18,440 eight-year-old evaluated.  In the state of Utah, a fairly rural state, there are 45 eight-year-old diagnosed with ASD out of 2,123 eight years old evaluated ("Autism and developmental," 2008).  Is there enough evidence to show that the proportion of children diagnosed with ASD in Pennsylvania is more than the proportion in Utah?  Test at the 1% level.

Suppose that  P1 is the proportion of children diagnosed for Pennsylvania and P2 is the proportion of children diagnosed for Utah.

Null hypothesis H0: P1 = P2

Alternate Hypothesis: H1: P1 > P2 (claim)

Now given that, level of significance is 0.01

Critical z for right tailed test is 2.33.

Hence, rejection region is z > 2.33

Pooled proportion = (245 + 45)/ (18440 + 2123) = 0.0141

Standard error of proportion = 0.0141(1-0.0141)(118440+12123) = 0.0027

Calculation of z test,

Z test = 24518440-4521230.0027 = -2.93

Z test is less than 2.33. So, it is not in the rejection region. Hence, we fail to reject the null hypothesis. Hence, insufficient evidence that the proportion of children diagnosed with ASD in Pennsylvania is more than the proportion in Utah.

 

9.2.3

All Fresh Seafood is a wholesale fish company based on the east coast of the U.S.  Catalina Offshore Products is a wholesale fish company based on the west coast of the U.S.  Table #9.2.5 contains prices from both companies for specific fish types ("Seafood online," 2013) ("Buy sushi grade," 2013).  Do the data provide enough evidence to show that a west coast fish wholesaler is more expensive than an east coast wholesaler?  Test at the 5% level.

Table #9.2.5: Wholesale Prices of Fish in Dollars

 

 

Fish

All Fresh Seafood Prices

Catalina Offshore Products Prices

Cod

19.99

17.99

Tilapi

6.00

13.99

Farmed Salmon

19.99

22.99

Organic Salmon

24.99

24.99

Grouper Fillet

29.99

19.99

Tuna

28.99

31.99

Swordfish

23.99

23.99

Sea Bass

32.99

23.99

Striped Bass

29.99

14.99

 

Suppose that µd = Mean of paired differences for east cost – west cost

Null Hypothesis: H0: µd = 0

Alternate Hypothesis: H1: µd < 0>

Given that, level of significance is 0.05

Degree of freedom DF = 9 – 1 = 8

Critical t for the left tailed test is -1.86

Rejection region is  t < -1.86.

I used a data analysis tool in excel to perform paired t-test:

t-Test: Paired Two Sample for Means

       

 

All Fresh Seafood Prices

Catalina Offshore Products Prices

 

Mean

24.10222

21.65667

 

Variance

66.81584

31.25

 

Observations

9

9

 

Pearson Correlation

0.473953

   

Hypothesized Mean Difference

0

   

df

8

   

t Stat

0.991517

   

P(T<=t) one-tail

0.175236

   

t Critical one-tail

1.859548

   

P(T<=t) two-tail

0.350472

   

t Critical two-tail

2.306004

 

 

 

T test = 0.992

Here, t-test is not in the rejection region. So, we cannot reject the null hypothesis. So, insufficient evidence to support the claim that a west coast fish wholesaler is more expensive than an east coast wholesaler.

 

 

 

 

 

9.2.6

The British Department of Transportation studied to see if people avoid driving on Friday the 13th.   They did a traffic count on a Friday and then again on a Friday the 13th at the same two locations ("Friday the 13th," 2013).  The data for each location on the two different dates is in table #9.2.6.  Estimate the mean difference in traffic count between the 6th and the 13th using a 90% level.

Table #9.2.6: Traffic Count

Dates

6th

13th

1990, July

139246

138548

1990, July

134012

132908

1991, September

137055

136018

1991, September

133732

131843

1991, December

123552

121641

1991, December

121139

118723

1992, March

128293

125532

1992, March

124631

120249

1992, November

124609

122770

1992, November

117584

117263

 

Calculation of paired differences:

Dates

6th

13th

d

1990, July

139246

138548

698

1990, July

134012

132908

1104

1991, September

137055

136018

1037

1991, September

133732

131843

1889

1991, December

123552

121641

1911

1991, December

121139

118723

2416

1992, March

128293

125532

2761

1992, March

124631

120249

4382

1992, November

124609

122770

1839

1992, November

117584

117263

321

 

Now using excel calculator,

Mean of differences µd = 1835.8

Standard deviation s = 1176.014

Critical t for DF 9 and confidence level 90% = 1.833

So, margin of error for confidence interval calculation is E = 1.833*1176.014/SQRT(10) = 681.67

90% confidence interval = (1835.8 – 681.67, 1835.8 + 681.67) = (1154.13, 2517.47)

We, can be 90% confident that true mean difference in traffic count between the 6th and the 13th will lie within (1154.13, 2517.47)

9.3.1

The income of males in each state of the United States, including the District of Columbia and Puerto Rico, are given in table #9.3.3, and the income of females is given in table #9.3.4 ("Median income of," 2013).  Is there enough evidence to show that the mean income of males is more than of females?  Test at the 1% level.

 

Table #9.3.3: Data of Income for Males

$42,951

$52,379

$42,544

$37,488

$49,281

$50,987

$60,705

$50,411

$66,760

$40,951

$43,902

$45,494

$41,528

$50,746

$45,183

$43,624

$43,993

$41,612

$46,313

$43,944

$56,708

$60,264

$50,053

$50,580

$40,202

$43,146

$41,635

$42,182

$41,803

$53,033

$60,568

$41,037

$50,388

$41,950

$44,660

$46,176

$41,420

$45,976

$47,956

$22,529

$48,842

$41,464

$40,285

$41,309

$43,160

$47,573

$44,057

$52,805

$53,046

$42,125

$46,214

$51,630

 

 

 

 

 

Table #9.3.4: Data of Income for Females

$31,862

$40,550

$36,048

$30,752

$41,817

$40,236

$47,476

$40,500

$60,332

$33,823

$35,438

$37,242

$31,238

$39,150

$34,023

$33,745

$33,269

$32,684

$31,844

$34,599

$48,748

$46,185

$36,931

$40,416

$29,548

$33,865

$31,067

$33,424

$35,484

$41,021

$47,155

$32,316

$42,113

$33,459

$32,462

$35,746

$31,274

$36,027

$37,089

$22,117

$41,412

$31,330

$31,329

$33,184

$35,301

$32,843

$38,177

$40,969

$40,993

$29,688

$35,890

$34,381

 

 

 

 

 

Suppose that mean income for male is µ1 and mean income for female is µ2

Null Hypothesis: H0: µ1 = µ2

Alternate Hypothesis: H1: µ1 > µ2 (claim)

Given that Level of significance ? = 0.01

I used excel to perform independent sample t test.

t-Test: Two-Sample Assuming Unequal Variances

     

 

Males

Females

Mean

46446.38

36511

Variance

49473354

37676539

Observations

52

52

Hypothesized Mean Difference

0

 

df

100

 

t Stat

7.67455

 

P(T<=t) one-tail

5.65E-12

 

t Critical one-tail

2.364217

 

P(T<=t) two-tail

1.13E-11

 

t Critical two-tail

2.625891

 

 

T test = 7.675

P = 0

As P < level xss=removed>the mean income of males is more than of females.

9.3.3

A study was conducted that measured the total brain volume (TVB) (in ) of patients that had schizophrenia and patients that are considered normal.  Table #9.3.5 contains the TVB of the normal patients and table #9.3.6 contains the TVB of schizophrenia patients ("SOCR data oct2009," 2013).  Is there enough evidence to show that the patients with schizophrenia have less TBV on average than a patient that is considered normal?  Test at the 10% level.

 

Table #9.3.5: Total Brain Volume (in ) of Normal Patients

1663407

1583940

1299470

1535137

1431890

1578698

1453510

1650348

1288971

1366346

1326402

1503005

1474790

1317156

1441045

1463498

1650207

1523045

1441636

1432033

1420416

1480171

1360810

1410213

1574808

1502702

1203344

1319737

1688990

1292641

1512571

1635918

 

 

 

 

 

Table #9.3.6: Total Brain Volume (in ) of Schizophrenia Patients

1331777

1487886

1066075

1297327

1499983

1861991

1368378

1476891

1443775

1337827

1658258

1588132

1690182

1569413

1177002

1387893

1483763

1688950

1563593

1317885

1420249

1363859

1238979

1286638

1325525

1588573

1476254

1648209

1354054

1354649

1636119

 

 

 

 

 

 

Suppose that mean volume for normal patients is µ1 and mean volume for Schizophrenia patient is µ2

Null Hypothesis: H0: µ1 = µ2

Alternate Hypothesis: H1: µ1 > µ2 (claim)

I used excel to perform independent sample t test.

t-Test: Two-Sample Assuming Unequal Variances

     

 

Normal

Schizophrenia

Mean

1463339

1451293

Variance

1.57E+10

2.96E+10

Observations

32

31

Hypothesized Mean Difference

0

 

df

55

 

t Stat

0.316843

 

P(T<=t) one-tail

0.376281

 

t Critical one-tail

1.297134

 

P(T<=t) two-tail

0.752562

 

t Critical two-tail

1.673034

 

 

P = 0.3168

As P > level of significance 0.10, we fail to reject the null hypothesis.

Hence insufficient evidence to support the claim that the patients with schizophrenia have less TBV on average than a patient that is considered normal.

 

9.3.4

A study was conducted that measured the total brain volume (TBV) (in ) of patients that had schizophrenia and patients that are considered normal.  Table #9.3.5 contains the TBV of the normal patients and table #9.3.6 contains the TBV of schizophrenia patients ("SOCR data oct2009," 2013).  Compute a 90% confidence interval for the difference in TBV of normal patients and patients with Schizophrenia.

 

Suppose that mean volume for normal patients is µ1 and mean volume for Schizophrenia patient is µ2

Output for two sample t test confidence interval calculator:

Two sample T confidence interval:


?1 : Mean of Normal
?2 : Mean of Schizophrenia
?1 - ?2 : Difference between two means
(without pooled variances)

90% confidence interval results:

Difference

Sample Diff.

Std. Err.

DF

L. Limit

U. Limit

?1 - ?2

12046.025

38018.926

54.816618

-51564.575

75656.626

 

90% confidence interval = (-51564.6, 75656.6)

Hence, we can be 90% confident that the true mean difference in TVB of normal patients and patients with Schizophrenia will lie within (-51564.6, 75656.6)

As 0 is within the confidence interval, there is no any significant difference between TBV of normal patients and patients with Schizophrenia.

 

 

9.3.8

The number of cell phones per 100 residents in countries in Europe is given in table #9.3.9 for the year 2010.  The number of cell phones per 100 residents in countries of the Americas is given in table #9.3.10 also for the year 2010 ("Population reference bureau," 2013).  Find the 98% confidence interval for the difference in a mean number of cell phones per 100 residents in Europe and the Americas.

 

Table #9.3.9: Number of Cell Phones per 100 Residents in Europe

100

76

100

130

75

84

112

84

138

133

118

134

126

188

129

93

64

128

124

122

109

121

127

152

96

63

99

95

151

147

123

95

67

67

118

125

110

115

140

115

141

77

98

102

102

112

118

118

54

23

121

126

47

 

 

Table #9.3.10: Number of Cell Phones per 100 Residents in the Americas

158

117

106

159

53

50

78

66

88

92

42

3

150

72

86

113

50

58

70

109

37

32

85

101

75

69

55

115

95

73

86

157

100

119

81

113

87

105

96

 

 

 

 

Output for two-sample t-test confidence interval calculator:

 

Two sample T confidence interval:


?1 : Mean of Europe
?2 : Mean of America
?1 - ?2 : Difference between two means
(without pooled variances)

98% confidence interval results:

Difference

Sample Diff.

Std. Err.

DF

L. Limit

U. Limit

?1 - ?2

20.945815

6.9736187

74.029011

4.3640739

37.527556

 

98% confidence interval = (4.364, 37.528)

Hence, we can be 98% confident that true mean difference in mean number of cell phones per 100 residents in Europe and the Americas.

As the confidence interval don’t include 0, there is a significant difference between mean number of cell phones per 100 residents in Europe and the Americas.

Mean number of cell phones per 100 residents in Europe is significantly more than Americans.

11.3.2

Levi-Strauss Co manufactures clothing.  The quality control department measures weekly values of different suppliers for the percentage difference of waste between the layout on the computer and the actual waste when the clothing is made (called run-up).  The data is in table #11.3.3, and there are some negative values because sometimes the supplier is able to layout the pattern better than the computer ("Waste run up," 2013).  Do the data show that there is a difference between some of the suppliers?  Test at the 1% level.

Table #11.3.3: Run-ups for Different Plants Making Levi Strauss Clothing

Plant 1

Plant 2

Plant 3

Plant 4

Plant 5

1.2

16.4

12.1

11.5

24

10.1

-6

9.7

10.2

-3.7

-2

-11.6

7.4

3.8

8.2

1.5

-1.3

-2.1

8.3

9.2

-3

4

10.1

6.6

-9.3

-0.7

17

4.7

10.2

8

3.2

3.8

4.6

8.8

15.8

2.7

4.3

3.9

2.7

22.3

-3.2

10.4

3.6

5.1

3.1

-1.7

4.2

9.6

11.2

16.8

2.4

8.5

9.8

5.9

11.3

0.3

6.3

6.5

13

12.3

3.5

9

5.7

6.8

16.9

-0.8

7.1

5.1

14.5

 

19.4

4.3

3.4

5.2

 

2.8

19.7

-0.8

7.3

 

13

3

-3.9

7.1

 

42.7

7.6

0.9

3.4

 

1.4

70.2

1.5

0.7

 

3

8.5

 

 

 

2.4

6

 

 

 

1.3

2.9

 

 

 

 

 

 

Null Hypothesis: Mean Run-ups for all the 5 plants are equal to each other.

Alternate Hypothesis: Mean Run-ups for at least one plant is different from others.

Level of significance ? = 0.01

Excel output for one way ANOVA:

Anova: Single Factor

         
             

SUMMARY

         

Groups

Count

Sum

Average

Variance

   

Plant 1

22

99.5

4.522727

100.6418

   

Plant 2

22

194.3

8.831818

235.7289

   

Plant 3

19

91.8

4.831579

19.38784

   

Plant 4

19

142.3

7.489474

13.37433

   

Plant 5

13

134.9

10.37692

91.29859

   
             
             

ANOVA

           

Source of Variation

SS

df

MS

F

P-value

F crit

Between Groups

450.9207

4

112.7302

1.159631

0.334012

3.534992

Within Groups

8749.088

90

97.21209

     
             

Total

9200.009

94

 

 

 

 

 

P = 0.334

As P > level of significance 0.01, we fail to reject the null hypothesis.

So insufficient evidence to support the claim that there is a difference between Run-ups for some of the suppliers.

 

 

11.3.4

A study was undertaken to see how accurate food labeling for calories on food that is considered reduced-calorie.  The group measured the number of calories for each item of food and then found the percent difference between measured and labeled food, .  The group also looked at food that was nationally advertised, regionally distributed, or locally prepared.  The data is in table #11.3.5 ("Calories datafile," 2013).  Do the data indicate that at least two of the mean percent differences between the three groups are different?  Test at the 10% level.

Table #11.3.5: Percent Differences Between Measured and Labeled Food

National Advertised

Regionally Distributed

Locally Prepared

2

41

15

-28

46

60

-6

2

250

8

25

145

6

39

6

-1

16.5

80

10

17

95

13

28

3

15

-3

 

-4

14

 

-4

34

 

-18

42

 

10

 

 

5

 

 

3

 

 

-7

 

 

3

 

 

-0.5

 

 

-10

 

 

6

 

 

 

Null Hypothesis: Mean Percent Differences Between Measured and Labeled Food for all the 3 groups are equal to each other.

Alternate Hypothesis: Mean Percent Differences Between Measured and Labeled Food for at least one group is different from others.

Level of significance = 0.10

 

 

Excel output for data tools one way ANOVA:

Anova: Single Factor

         
             

SUMMARY

         

Groups

Count

Sum

Average

Variance

   

National Advertised

20

2.5

0.125

110.6809

   

Regionally Distributed

12

301.5

25.125

258.3693

   

Locally Prepared

8

654

81.75

7050.786

   
             
             

ANOVA

           

Source of Variation

SS

df

MS

F

P-value

F crit

Between Groups

38095.9

2

19047.95

12.97915

5.36E-05

2.452014

Within Groups

54300.5

37

1467.581

     
             

Total

92396.4

39

 

 

 

 

 

P = 0

As P < level>

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