# 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

## 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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