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 µ_d = Mean of paired differences for east cost – west cost

Null Hypothesis: H₀: µ_d = 0

Alternate Hypothesis: H₁: µ_d < 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 13^th.   They did a traffic count on a Friday and then again on a Friday the 13^th 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 6^th and the 13^th 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 µ_d = 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 6^th and the 13^th 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 thedifference 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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