Practice Problems for Exam III 1) You wish to investigate whether gender and salary levels are independent. A random sample of 140 adults is selected and the results are classified in the following table. At 1% level of significance, is there sufficient evidence to conclude that gender and salary are dependent? $10,000- $24,999 $25,000- $49,999 Over $50,000 Male 5 25 50 Female 5 40 15 2) A math professor claims that the standard deviation of IQ scores of the students majoring in mathematics is different from 15, where 15 is the standard deviation of IQ scores for the general population. To test his claim, 10 students were surveyed and their IQ scores were recorded. 120, 110, 113, 123, 119, 160, 220, 119, 110, 140 Test the professor’s claim using a 0.01 level of significance. 3) The table below shows the number of hours per day a sample of patients suffered from headaches before and after seven weeks of receiving a new drug. At 1% level of significance, test the claim that the new drug helps to reduce the number of daily headache hours. Before 4.1 3.6 2.5 4.0 1.3 2.4 1.1 After 3.8 3.7 2.7 2.5 1.4 1.6 0.6 4) In crash tests at 5 miles per hour, the mean bumper repair cost for 14 small cars is $560 with a standard deviation of $185. In similar tests of 23 midsize cars, the mean bumper repair cost is $620 with a standard deviation of $170. At 1% significance level, test the claim that the mean bumper repair cost is less for small cars than it is for midsize cars. 5) A car magazine is claims that the standard deviation of the repair costs incurred during the first three years on the Acura TL is less than $200. The repair costs in dollars for randomly selected eight Acura are show below: 420 810 130 0 400 533 120 459 Test the claim using a 0.01 level of significance. 6) Shoe Size Height A 8 65 B 9 70 C 7 60 D 9 75 a) You select four men and measure their heights (in inches) and shoe size of each. Use a 0.01 level of significance to test the claim that there is a correlation between the men’s shoe sizes and heights. b) Use the regression equation to predict the height of a man with shoe size 8.
Transcript
Practice Problems for Exam III 1) You wish to investigate whether gender and salary levels are independent. A random sample
of 140 adults is selected and the results are classified in the following table. At 1% level of significance, is there sufficient evidence to conclude that gender and salary are dependent?
$10,000-$24,999
$25,000-$49,999
Over $50,000
Male 5 25 50Female 5 40 15
2) A math professor claims that the standard deviation of IQ scores of the students majoring in mathematics is different from 15, where 15 is the standard deviation of IQ scores for the general population. To test his claim, 10 students were surveyed and their IQ scores were recorded.
120, 110, 113, 123, 119, 160, 220, 119, 110, 140 Test the professor’s claim using a 0.01 level of significance.
3) The table below shows the number of hours per day a sample of patients suffered from headaches before and after seven weeks of receiving a new drug. At 1% level of significance, test the claim that the new drug helps to reduce the number of daily headache hours.
4) In crash tests at 5 miles per hour, the mean bumper repair cost for 14 small cars is $560 with a standard deviation of $185. In similar tests of 23 midsize cars, the mean bumper repair cost is $620 with a standard deviation of $170. At 1% significance level, test the claim that the mean bumper repair cost is less for small cars than it is for midsize cars.
5) A car magazine is claims that the standard deviation of the repair costs incurred during the first three years on the Acura TL is less than $200. The repair costs in dollars for randomly selected eight Acura are show below:
420 810 130 0 400 533 120 459 Test the claim using a 0.01 level of significance.
6)
ShoeSize
Height
A 8 65B 9 70C 7 60D 9 75
a) You select four men and measure their heights (in inches) and shoe size of each. Use a 0.01 level of significance to test the claim that there is a correlation between the men’s shoe sizes and heights.
b) Use the regression equation to predict the height of a man with shoe size 8.
7) On April 4-11, 2015, Gallup surveyed 535 adults and asked “Do you think there is life of some form on the other planets in the universe or not?” Of the 535 adults surveyed, 310 responded “Yes”. When the same questions were asked on March 1-8, 2001 , 349 of the 535 individuals surveyed responded “Yes”. Test the claim that the proportion of adults who believe that there is a life on other planets in 2015 has decreased since 2001. Use 0.05 level of significance.
8) A math instructor claims that college women have more credit card debt than college men. He conducts a random sample of 8 college women and 8 college men and obtains the following results (in dollars) :Women: 2000 280 180 2100 310 4200 120 230
Men: 120 200 1200 3200 130 310 240 520 Test the claim at a 0.05 significance level You may assume that credit card debts are normally distributed.
9) In a survey of 3420 students attending private high schools, 910 said they had smoked in the last 30 days. In a survey of 5100 college students attending public high schools, 1500 said they had smoked in the last 30 days. At a 0.01 significance level, can you support the claim that the proportion of high school students who said they had smoked in the last 30 days in the private schools is different from the proportion in public schools?
10)A study was conducted to investigate a relationship between self-efficacy scores and their SAT scores. 2a) Is there a sufficient evidence to indicate that self-efficacy scores and SAT scores are
positively correlated? Test the claim at 1% level.b) Find the regression equation and predict the SAT score if a student received a 3 on the self-
11) A sixth grade teacher in Walnut gave a math test to a random sample of 36 students. The mean and standard deviation for are 73.12 and 8.91 respectively. The teacher claims that the mean performance for her students is higher than the national mean of 72. At the 1% significance level, do the data support her claim?
12) In a study conducted by a Mt. SAC researcher, study participants were asked to react to a stimulus. In one experiment, the participant must press a key on seeing blue dots. Reaction time to press the key (in seconds) is measured. The same person is then asked to press a key on seeing red dots, again with reaction time measured. Test the claim that the reaction time to the blue stimulus is different from the reaction time to the red stimulus. Use 0.05 for the significance level.
Reaction Time to Blue
4.3 1.4 3.5 2.0 3.2
Reaction Time to Red
2.4 1.6 2.4 2.8 3.1
13) The National Center for Educational Statistics publishes the results from the trends in international Math and Science Study. The table below contains the 2000 and 2010 scores. Use 0.01 level of significance, test the claim that the scores have improved from 2000 to 2010.
14)A recent Gallup Poll asked male and female Americans whether they were pro life or pro choice. Test whether individual’s opinion regarding abortion is independent of gender using 0.05 level of significance.
Pro Life Pro ChoiceMale 5 25Female 23 40
15)An automotive battery manufacture claims that the mean life of their batteries is different from 60 months. A sample of 50 batteries had a mean life of 54.12 months and a standard deviation of 3.2 months. Test the manufacture’s claim at the 1% significance level.
16)HDL is known as “good cholesterol” since high levels of HDL lower the risk of heart diseases. A nutritionist claims that male runners have higher HDL levels than male non-runners. The data below list HDL levels of runners and non-runners. Test the claim using a 0.05 level of significance.
17)A store manager claims that the standard deviation of the number of customers per day is more than 8. A random sample of 51 days has a mean number of customers 498 with a standard deviation 8.2. Test the claim using a 0.05 level of significance.
18)A recent study found that the average age of robbery victims is 64.2 years old. A Mt. SAC sociologist claims that the mean age of robbery victims in Walnut is different from 64.2 years old. The data below give the age of 10 recent robbery victims in Walnut. Test the claim using a 0.01 level of significance.
56 65 54 43 19 31 77 73 47 51
19)A large national study found that 23.1% of the US children speak a language other than English at home. A Mt.SAC sociologist claims that the percentage is higher for the children in Walnut. A random sample of 440 children in Walnut found 35% spoke a language other than English at home. Test the claim using a 0.05 level of significance.
20)A nutritionist claims that there is a correlation between the number of calories and the number of cholesterol for fast food hamburgers. The number of calories and the number of cholesterol for hamburgers from seven fast food restaurants are shown below. Use a 0.05 level of significance to test the claim.
Calories 430 544 436 563 532 567 432
Cholesterol 43 32 35 34 43 46 32
21) In a random sample of 400 men, 42% indicated that they wished they were rich. In a random sample of 600 women, 48% indicated that they wished they were rich. Is there a significant difference in the proportions? Use a 0.01 level of significance.
22)A sociologist in Walnut claims that the mean salary of people with college degrees is higher than the mean salary of people without college degrees. The data below list salaries of people with college degrees and without college degrees. Test the claim using a 0.05 level of significance.
College degrees $40,000 $67,000 $90,000 $76,000 $20,000Without college degrees $97,000 $56,000 $80,000 $30,000 $19,000 $18,000
23)The U.S. Census Bureau tracks trends in minority ownership of businesses. A random sample of 1000 businesses in Michigan showed 107 were minority owned. Another random sample of 1000 businesses in Minnesota showed 87 were minority owned. Using a 0.05 level of significance, test the claim that the population proportion of minority owned businesses in Michigan is higher than that of Minnesota.
24)The U.S. Census Bureau reported that the mean age at first marriage for females in 2002 was 25.3. A random sample of 42 females in 2012 yielded a mean age at first marriage of 28.1 with a standard deviation of 3.1 years. At a 0.01 level of significance, test the claim that the population mean age at first marriage for females has increased since 2002.
25)An advertisement for a phonetics game claims that it will improve reading scores. The reading levels of eight randomly selected children were tested without the phonetics games and retested after exposure to the game. Test the claim using a 0.01 level of significance.
26)USA Today reported that 47% of the general consumer population in the US is loyal to the automotive manufacturer of their choice. Toyota conducted a study of a random sample of 900 Toyota owners and found that 480 of them said they would buy another Toyota. Test the claim that the population proportion of consumers loyal to Toyota is greater than 47%. Use a 0.01 significance level.
Traditional P-value
27)The director of an alumni association for Mt. SAC claim that there is a negative correlation between the amount of an alumnus’s contribution and the years the alumnus has been out of school. Use a 0.01 level of significance to test the claim
Years 13 31 9 7 11 8 2 9Contribution (in dollars)
1000 10000 8000 1500 20000 1000 100 5000
28)The National Institute on Alcohol Abuse and Alcoholism reported in 2009 that 45.9% of eighth graders had used alcohol. A random sample of 100 eighth graders this year showed 43% of them used alcohol. Test the claim using a 0.01 level of significance that the population proportion of eighth graders who used alcohol has changed since 2009.
Traditional P-value
29)A educator claims that the number hours spent on homework and final grade are positively correlated. The data obtained from a random sample of seven students are shown below. Test the claim using a 0.05 level of significance.
# of HW hours
13 34 17 41 10 20 10
Final grade 78 87 56 90 48 80 13
Traditional P-value
30)A study compared the IQ scores of children with autism and children with Asperger’s syndrome. A random sample of 28 children with autism had a mean IQ of 90 and a standard deviation of 13. A random sample of 32 children with Asperger’s syndrome had a mean IQ of 85 and a standard deviation of 19. Using a 0.05 level of significance, test the claim that the mean IQ of children with autism is higher than the mean IQ of children with Asperger’s syndrome.
31)A survey conducted in 2001 found that 20% of Mt. SAC students are Baptists. A sociologist claims that the percentage has changed since 2001. A survey of 500 randomly selected Mt. SAC students showed that 19% were Baptists. Test the claim using a 0.03 level of significance.
Traditional P-value
32)A tobacco company advertises that only 18% of Americans are in favor of outlawing cigarettes. A Mt. SAC student claims that more than 18% of Mt. SAC students are in favor of outlawing cigarettes. In a random sample of 1205 Mt. SAC students, she finds that 341 are in favor of outlawing cigarettes. Test the student’s claim at 3% significance level.
33) (3 points) Find the regression equation and predict the income for someone with 13 years of education.
# of years of education
13 11 19 3 18 14 14 13
Income $31,000 $28,000 $61,000 $25,000 $82,000 $76,000 $43,000 $53,000
34)a. Give an example of two variables that are positively correlated.b. Give an example of two variables that are negatively correlated.c. Sketch a scatter diagram using 4 ordered pairs whose correlation coefficient is -1.
35)Determine if the following variables are positively correlated , negatively correlated, or not correlated. Justify your answer
a. SAT score and college GPAb. Income and risk of cancer
36)
Answer Key (Again, I have brain damage )1) Traditional P-valueH0 :salary and gender are independentH1 :salary and gender are dependent
H0 :salary and gender are independentH1 :salary and gender are dependent
Store the matrix: matrix>Edit>enter the dimensions of the matrix.
Then stat>tests>χ2−test
2) χ2=19/856 2) P-value =4.88 x 10-5
3) Use the χ2
-table, always right tailed, df=2 is given in the TI output
3)P−value≤α . Reject H0
4) Reject H0 The findings suggest that gender and salary are dependent.
5) Possible Type I Error
2)Traditional P-value
H0 : σ=15H1 :σ≠15
Do not use TI: use χ2=
(n−1)s2
σ2, where σ is obtained from H0
2)χ2=
9(34 .1)2
152 =46 .5
3) Go to the χ2− table, use
1−α2 and
α2 )
χ2=1 .735( left ) and 23 ,589(right ).
4) Reject H0 The findings suggest that the standard deviation of IQ scores of the students majoring in mathematics is different from 15
5) Possible Type I error
3)
Traditional P-valueH0 : μd=0H1 : μd>0
H0 : μd=0H1 : μd>0
Compute L1 –L2 = L3: Store the values in L1 and L2 by Stat>Edit. Then L1-L2 >STO>L3
Use Stat>Tests> T-test; Make sure μ0=0 , use List L32) t=1.667 3) p=0.0732
3) For the critical region, since this is a right tailed test, first compute 1-0.01=0.99, df = 7-1= 6 Use invT(0.99,6)=3.143)
)P−value>α . FTR H0
4) ) Fail to reject H0 The findings fail to suggest that the new drug helps to reduce the number of daily headache hours
5) Possible TYPE II
4)
Traditional P-valueH0 : μ1=μ2H1 : μ1<μ2
H0 : μ1=μ2H1 : μ1<μ2
Stat>Tests>2-sample t test Choose Stats2) t= -.986 P-value= .166
4) Fail to reject H0 The findings fail to suggest that the mean bumper repair cost is less for small cars than it is for midsize cars
6) Possible Type II error
5)
Traditional P-valueH0 : σ=200H1 :σ<200First compute the sample standard deviation S by Stat>Edit>L1 Then Stat>Calc>one var stats
2)Use the formula χ2=
(n−1)s2
σ2, where σ is
obtained from H0
χ2=7(263 )2
2002 =12. 11
3)Go to the table, use 1- 0.01=0.99χ2=1 .239
4) 4) ) Fail to reject H0 The findings fail to suggest that the standard deviation of the repair costs is less than $200
6) Possible Type II error
6)
a)Traditional P-valueH0 : ρ=0H1 : ρ≠0
H0 : ρ=0H1 : ρ≠0
First store the values in L1 and L2: Stat>Edit. Then Stat>Tests>LINREGTTEST (linear regression t-test)2) t=4.041 2) P=0.56123) C.V. = -9.925, 9.925 (since this is a 2-talied test, divide the level of significant by 2, obtaining 0.005. Then use invT(0.005, 2)=-9.925.)
)P−value>α . FTRH0
4) Fail to reject H0 The findings fail to suggest that there is a correlation between the men’s shoe sizes and heights.
6) Possible Type II error
b) y=a+bx=15+6. 4 x . When x is 8, y=15+6 . 4(8 )=66 .2 inches is the predicted height.
Reject H0 The findings suggest that the proportion of adults who believe that there is a life on other planets in 2013 has decreased since 2001Possible Type I error
8)
Traditional P-valueH0 : μ1=μ2H1 : μ1>μ2
H0 : μ1=μ2H1 : μ1>μ2
First store the values in L1 and L2. Stat>Edit. Then Stat>Tests>2-sapleT testt = .682 , P-value= .2537c.v.= 1.895 (Since this is a right-tailed test, compute 1-0.05=0.95. df=8-1=7 Then use invT(0.95,7)=1.895)
P−value>α . FTR H0
4) Fail to Reject H0 The findings fail to suggest that the mean credit card debt of college women is higher than the mean credit card debt of college men.
5) Possible Type I error
9)
Traditional P-valueH0 : p1=p2H1 : p1≠p2
H0 : p1=p2H1 : p1≠p2
Stat>Tests>2-prop z-testz = -2.82 P-value = .004863) cv = 2.576, -2.576 (since this is a 2-tailed test, divide the significance level by 2. Then use invNorm(0.005, 0, 1)=-2.58
P−value≤α . Reject H0
d) Reject H0 The findings suggest that the proportion of high school students who said they
had smoked in the last 30 days in the private schools is different from the proportion in public schools5) Possible Type I error
10)
Traditional P-valueH0 : ρ=0H1 : ρ>0
H0 : ρ=0H1 : ρ>0
First Store the values in L1 and L2: Stat>Edit. Then Stat>Tests>Linear Reg Testt=.7090 P-value= .2279C.V. = 3.365 (Since this is a right-tailed test, compute 1-0.01=0.99. df== 7 – 2 = 5 Then use invT(0.99, 5)) ,
P−value>α . FTR H0
) Fail to reject H0 The findings fail to suggest that self-efficacy scores and SAT scores are positively correlated.5) Type II
b) y=552+22 . 7x , when x = 3, the SAT math score is predicted to be 552+22.7(3)= 620.1
11)
Traditional P-valueH0 : μ=72H1 : μ>72
H0 : μ=72H1 : μ>72
Stat>tests>t test Choose Statst=0 . 754 P-value= .2279
) critical value: t =2 .441(Since this is a right-tailed test, compute 1-0.01=0.99. Then use invT(0.99, 35))
)P−value>α . FTR H0
4) Fail to reject H0 The findings fail to suggest that the mean performance for her students is higher than the national mean of 72.55) Possible Type II error
12)
Traditional P-value
H0 : μd=0H1 : μd≠0
H0 : μd=0H1 : μd≠0
First store the values in L1 and L2: stat>edit. Then Compute L1 –L2 = L3 : L1-L2 stoL3
Then Stat>tests>t test Make sure μ0=0 , use List L3t-test t=.8731 P-value=4310: use invT(0.01/2, 5))=-4.032 P−value>α . FTR H0
4) FTR H0 The findings fail to suggest that the reaction time to the blue stimulus is different from the reaction time to the red stimulus. 5) Type II
13) )
Traditional P-valueH0 : μd=0H1 : μd<0
H0 : μd=0H1 : μd<0
First store the values in L1 and L2. Then ) Compute L1 –L2 = L3 Then Stat>tests>t test Make sure μ0=0 , use List L3t-test t=1.000, , P-value = .3631
CV: t = -3.365 =invT(0.01, 5) P−value>α . FTR H0
FTR H0 The findings fail to suggest that the scores have improved from 2000 to 20105) Type II
14) Traditional P-valueH0 :opinions and gender are independentH1 :opinions and gender are dependent
H0 :opinions and gender are independentH1 :opinions and gender are dependent
Store the matrix: matrix>Edit>enter the dimensions of the matrix.
Then stat>tests>χ2−test
χ2=3 . 802 P = 0.0512
) c.v. χ2=3 .841(from the table: remember, the
contingency table test is always right-tailed) ,P−value>α . FTR H0
Fail to reject H0 The findings fail to suggest that individual’s opinion regarding abortion is dependent of gender.5) type II
15)
Traditional P-valueH0 : μ=60H1 : μ≠60
H0 : μ=60H1 : μ≠60
Stat>tests>t test . Choose Stats option
t=−12 . 993 P-value- 1 .71×10−17
critical value ±2 .680 (First divide the significance level by 2 since this is a 2 tailed test. Them use invT(0.005, 49)
P−value≤α . Reject H0
Reject H0 The findings suggest that the mean life of their batteries is different from 60 months.Possible Type I
16) .
1)
H0 : μ1=μ2H1 : μ1>μ2
H0 : μ1=μ2H1 : μ1>μ2
First store the values in L1 and L2Stat>tests>2-sample tt-test
2) T=-.431 P-value=.3373) invT(1-0.05, 6)=1.943 or the t-table P−value>α→Fail to reject H0
4) Fail to reject H0 . The findings fail to suggest that male runners have higher HDL levels than male non-runners
Possible Type II error.
17)
1)
H0 : σ=8H1 :σ>8
2) S=8.2
χ2=8 .22 (50)
82 =52 .531
3) 67.505 from the chi-square table.
4) FTR H0 . The findings fail to suggest that the standard deviation of the number of hours spent studying by Mt. SAC students is more than 2 hours per week.
4) Reject H0 . The findings suggest that the proportion of all children who speak a language other than English at home is greater than 23.1%
Possible Type I error.
5) Possible Type I errorAssume 23.1% of children speak a language other than English. The p-value is the probability of finding 35% of children speak who a language other than English in a random sample of 440 children.
20)
1)
H0 : ρ=0H1 : ρ≠0
H0 : ρ=0H1 : ρ≠0
2) T= .427 (LlNREGTEST) P-value=.6773) invT(0.05/2, 5)=-2.571 or the t-
table. For correlation problems, df=n-2
P−value>α→Fail to reject H0
4) Fail to reject H0 . The findings fail to suggest that there is a correlation between the number of calories and the number of cholesterol for fast food hamburgers.
Possible Type II error.
5) Possible Type II error This is a cross-sectional study
21)
1)
H0 : p1=p2H1 : p1≠p2
H0 : p1=p2H1 : p1≠p2
You must first find x1 , x2 using the formula x=n pStat>tests>2-prop z-test
2) Z=-1.87 (2 prop z test) P-value=.0602
3) invNorm(0.005, 0,1) =-2.58 P−value>α→Fail to reject H0
4) Fail to reject H0 . The findings fail to suggest that there is a difference between the proportion of men and women who wished they were rich
Possible Type II error.
5) Possible Type II error This is a cross-sectional study
22)
1)
H0 : μ1=μ2H1 : μ1>μ2
H0 : μ1=μ2H1 : μ1>μ2
First store the values in L1 and L2 Stat>EditStat>tests>2-sample tt-test
2) T=-.464 P-value=.32703) invT(1-0.05, 4)=2.132 or the t-table
(use df=5-1, the group with a smaller sample size)
P−value>α→Fail to reject H0
4) Fail to reject H0 . The findings fail to suggest that the mean salary of people with college degrees is higher than the mean salary of the people without college degrees
Possible Type II error.
23)
1)
H0 : p1=p2H1 : p1> p2
H0 : p1=p2H1 : p1> p2
x1 , x2 are given: no need to use the formula x=n pStat>tests>2-prop z-test
2) Z=-1.51 P-value=.0654
3) invNorm(1-0.05, 0,1) =1.65 P−value>α→Fail to reject H0
4) Fail to reject H0 . The findings fail to suggest that the proportion of minority owned business is higher in Michigan than in Minnesota.
Possible Type II error.
5) Possible Type II error This is a cross-sectional study
24)
1)
H0 : μ=25 .3H1 : μ>25 . 3
H0 : μ=25 .3H1 : μ>25 . 3
There is only one group. The 2002 study was conducted by someone else.Stat>tests>t test
4) Reject H0 The findings suggest that the mean age at first marriage has increased since 2002
Possible Type I error.
5) Possible Type I error
25) Traditional P-valueH0 : μd=0H1 : μd<0
H0 : μd=0H1 : μd<0
Compute L1 –L2 = L3: Store the values in L1 and L2 by Stat>Edit. Then L1-L2 >STO>L3
Use Stat>Tests> T-test; Make sure μ0=0 , use List L32) t=-.733 3) p=.243
2) Use invT(0.01, 7) -2.998)P−value>α . FTR H0
3) ) Fail to reject H0 The findings fail to indicate that phonics games are effective in improving reading scores.
4) Possible TYPE II This is a prospective study
26)
1)
H0 : p=0 .. 47H1 : p>0 . 47
H0 : p=0 .. 47H1 : p>0 .47
Use 1-prop Z Test: x=480, n=9002) Z=3.80 P-value=0 .00007033) invnorm(0.99,0,1) = 2.33
P−value≤α→Reject H0
4) Reject H0: . The findings suggest that the population proportion of consumers loyal to Toyota is greater than 47%.
Possible Type I error.
P-value: Assume that 47% of the customers are loyal to Toyota. The p-value is the probability of finding 480 or more customers loyal to Toyota in a random sample of 900 customers.
27)
1)
H0 : ρ=0H1 : ρ<0
H0 : ρ=0H1 : ρ<0
Store the values in L1, L2 Stat>EditStat>Tests>Linregttest (linear regression t-test)
2) T= 1.001 P-value=0.8223) invT(0.01, 6)=-3.143 or the t-
table. For correlation problems, df=n-2
P−value>α→Fail to reject H0
4) Fail to reject H0 . The findings fail to suggest that there is a negative correlation between the amount of an alumnus’s contribution and the years the alumnus has been out of school.
Possible Type II error.
5) Possible Type II error This is a prospective study
28)
1)
H0 : p=0 .459H1 : p≠0.459
H0 : p=0 .459H1 : p≠0.459
Use one-prop z-testX=100(0.43)=432) Z=-0.58 P-value==. 5613) C.V. 0.01/2=0.005
invnor(0.005,0,1) = -2.58, P−value>α→FTR H0
4) FTR H0 . The findings fail to suggest that the population proportion of eighth graders who used alcohol has changed since 2009.
Possible Type II error.
5) Possible Type II error This is a prospective study
29)
Traditional P-valueH0 : ρ=0H1 : ρ>0
H0 : ρ=0H1 : ρ>0
First Store the values in L1 and L2: Stat>Edit. Then Stat>Tests>Linear Reg Testt=2.343 P-value= .033C.V. = 2.015 (Since this is a right-tailed test, compute 1-0.01=0.99. df== 7 – 2 = 5 Then use invT(0.95, 5)) ,
P−value≤α . Reject H0
Reject H0 The findings suggest that the number hours spent on homework and final grade are positively correlated.5) Possible Type I Error
30)
Traditional P-valueH0 : μ1=μ2H1 : μ1>μ2
H0 : μ1=μ2H1 : μ1>μ2
Stat>Tests>2-sampleT test Choose the STATS optiont = 1.201 P-value= .1174
c.v.; df=27 P−value>α . FTR H0
4) Fail to Reject H0 The findings fail to suggest that the mean IQ of children with autism is higher than the mean IQ of children with Asperger’s syndrome.
31)
1)
H0 : p=0 .. 20H1 : p≠0.20
H0 : p=0 .. 20H1 : p≠0.20
Use 1-prop z-test: you must first find x by x=n p=500(0 .19 )=952) Z=-56 P-value=0 .576
3)α2=0 . 03
2=0 .015
invnorm(0.015,0,1) =- 2.17P−value>α→Fail to reject H0
4) FTR H0: . The findings fail to suggest that the percentage Mt. SAC students who are Baptists has changed since 2001.
4) Reject H0 . The findings Possible Type I error.
suggest that more than 18% of Mt. SAC students are in favor of outlawing cigarettes.
P-value: Assume 18% of Mt. SAC students are smokers. The P-value is the probability that 341 or more are smokers in a random sample of 1205 Mt. SAC students.
33) use STAT>Tests>LinRegTTest. Since you are only finding the regression equation, the selection of left/right/two tails does not affect your answer.
y=7471+3230 x . When x=13, the predicted income is 7471+3230(13 )=$ 49 ,46134)
a) # of cigarettes smoked and risk of cancerb) # of hours spent playing games and your GPAc) Make sure the four points lie on a decreasing straight line.
35)a) Positive. SAT is a predictor of success in collegeb) Negative. People in higher income brackets have access to better health care. In
addition, people in higher income brackets smoke less than the people in lower income brackets
