Practice Midterm

Welcome to Pstat5E: Welcome to Pstat5E: Statistics with Economics and Statistics with Economics and

Business ApplicationsBusiness Applications

Yuedong Wang

Solution to Practice Final Exam

Practice Midterm

1. Each year, billions of dollars are spent at theme parks owned by Disney, Universal Studios, Sea World and others. A management consultant claims that 20% of trips include a theme park visit. A survey of 1233 randomly selected people who took trips revealed that 111 of them visited a theme park.

(i) Construct a 95% confidence interval for the proportion of trips that include a theme park visit.

(ii) Do these data support the consultant's claim?

Practice Midterm

Solution: (i) We have a binomial experiment with p=proportion of trips include a theme park visit

(ii) Since the interval does not contain the value .2 (20%), the consultant’s claim is not supported.

.1060) ,0740(.

0160.0900. 1233

)1233/1111)(1233/111(96.1

1233

111

n

qpzp

:for interval confidence 95%

α/2

p

.1060) ,0740(.

0160.0900. 1233

)1233/1111)(1233/111(96.1

1233

111

n

qpzp

:for interval confidence 95%

α/2

p

Practice Midterm

2. A mathematical proficiency test were given to randomly selected 13-year-old male and female students. The following tables gives the sample mean scores and standard deviations:

(i) Estimate the difference in mean scores between male student and female students and construct the 95% confidence interval.

(ii) Can you conclude that the mean scores are different for male and female students?

Male Students Female Students

Sample size 905 905

Sample mean 474.6 473.2

Sample Std Dev 192.5 153.4

Practice Midterm

Solution: (i) Denote μ1=mean score for male students, μ2=mean score for female students. The point estimate of the difference, μ1-μ2, is

Since both sample sizes are large,

(ii) Since the confidence interval contains zero, we would not conclude that the mean scores are different between male and female students.

)44.17 ,64.14( ,04.164.1)(

:-for interval confidence 95%

2

22

1

21

2/21

21

n

s

n

szxx

)44.17 ,64.14( ,04.164.1)(

:-for interval confidence 95%

2

22

1

21

2/21

21

n

s

n

szxx

4.12.4736.47421 xx 4.12.4736.47421 xx

Practice Midterm

3. The paper ``The association of marijuana use with outcome of pregnancy'' (Amer. J. Public Health, 1983, pp.1161-1164) reported the following data on incidence of major malfunctions among newborns both for mothers who were marijuana users and for mothers who did not use marijuana.

User Nonuser

Sample size 1,246 11,178

Number of major malfunctions 42 294

Practice Midterm

(i) Construct a 99% confidence interval for the difference between the incidence rate among all mothers who use marijuana and the incidence rate among all mothers who do not use marijuana.

(ii) Do these data indicate that the incidence rate is higher for mothers who use marijuana?

Practice Midterm

Solution: (i) Denote

p1= incidence rate among all mothers who use marijuana,

p2= incidence rate among all mothers who do not use marijuana.

Since both sample sizes are large,

Practice Midterm

(ii) Since the confidence interval contains zero, we would not conclude that the incidence rate is higher for mothers who use marijuana

.0212) (-.0064,

.0138.0074

11178

9737.0263.

1246

9663.0337.58.20263.0337.

ˆˆˆˆ)ˆˆ(

:for interval confidence 99%

0263.11178/294ˆ ,0337.1246/42ˆ

2

22

1

112/21

21

21

n

qp

n

qpzpp

pp

pp

.0212) (-.0064,

.0138.0074

11178

9737.0263.

1246

9663.0337.58.20263.0337.

ˆˆˆˆ)ˆˆ(

:for interval confidence 99%

0263.11178/294ˆ ,0337.1246/42ˆ

2

22

1

112/21

21

21

n

qp

n

qpzpp

pp

pp

Practice Midterm

4. A new program has been developed to enrich the kindergarten experience of children in preparation for the first grade. Pupils in each classroom are tested at the beginning of the school year (pretest) and again at the end of the school year (posttest). The following table gives the scores of 9 randomly selected students exposed to the new curriculum (high score=better performance).

Pupil 1 2 3 4 5 6 7 8 9 x=Pretest 9 6 14 12 9 8 12 8 11 y=Posttest 16 11 14 10 14 12 15 11 14

Practice Midterm

(i) Apply an appropriate test to decide at the 5% level if the new curriculum significantly increased pupil's performance. Follow five steps in the lecture note.(ii) Specify assumptions for the above test.(iii) Suppose that further study establishes that, in fact, the population mean score at the beginning is 12.4 and the mean score at the end of the year is12.3.Refer back to part (i). Did your analysis lead to a (a) Type I error; (b) Type II error; (c) Correct decision; (d) None of (a)-(c).Circle the correct response.(iv) Do you change your conclusion in (i) if =.01?

Practice Midterm

Solution:(i) Since pretest and posttest scores come as pairs for

each pupil, the method we would use is the paired-difference test. Denote

x=pretest score, y=posttest score, d=x-y,

μ1=mean pretest score, μ2=mean posttest score.

Pupil 1 2 3 4 5 6 7 8 9

x=Pretest 9 6 14 12 9 8 12 8 11

y=Posttest 16 11 14 10 14 12 15 11 14

d=x-y -7 -5 0 2 -5 -4 -3 -3 -3

Practice Midterm

)(increased 0:H

t)improvemen (no 0:H

21a

210

)(increased 0:H

t)improvemen (no 0:H

21a

210

8191

44.39/7131.2

1111.3

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7131.2s ,1111.3d 9,n d

ndf

ns

dt

d

8191

44.39/7131.2

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7131.2s ,1111.3d 9,n d

ndf

ns

dt

d

Practice Midterm

.005valuep

005.3.355)P(t

3.44)P(t-3.44)P(tvalue-

sided-one :value-

p

p

.005valuep

005.3.355)P(t

3.44)P(t-3.44)P(tvalue-

sided-one :value-

p

p

Decision: since the p-value is smaller than = .05, H0 is rejected.

Conclusion: there is strong evidence that the new curriculum increases performance on average.

Decision: since the p-value is smaller than = .05, H0 is rejected.

Conclusion: there is strong evidence that the new curriculum increases performance on average.

Practice Midterm

(ii) The differences, d=x-y, are independent for different

pupils and have the same normal distribution.

(iii) μ1=12.4, μ2=12.3, H0 is true. Since we rejected H0, so we committed a type I error. Circle (a).

(iv) Since p-value < .01, we still reject H0.

Practice Midterm

5. An automobile manufacture recommends that any purchaser of one of its new cars bring it in to a dealer for a 3000-mile checkup. The company wishes to know whether the true average mileage for initial servicing differs from 3000.

(i) A random sample of 20 recent purchasers resulted in a sample average mileage of 3108 and a sample standard deviation of 273 miles. Does the data suggest that true average mileage for this checkup is something other than the recommended value? Use α=.01 and follow five steps in the lecture note.

Practice Midterm

(ii) In (i), instead of 20, suppose that the manufacture selected 50 recent purchasers, and gets the same sample mean and standard deviation as in (i). Does the data suggest that true average mileage for this checkup is something other than the recommended value?

Use α=.01.

(iii) In (ii), what is the smallest significance level that you will reject the null hypothesis?

(iv) Specify assumptions for the tests in (i) and (ii).

Practice Midterm

Solution:

(i) Denote μ=true average mileage of cars brought to the dealer for 3000-mile checkups.

3000:H ,3000:H a0 3000:H ,3000:H a0

191201

769.120/273

30003108

/

3000*

273s ,3108x size), sample (small 20n

ndfns

xt

191201

769.120/273

30003108

/

3000*

273s ,3108x size), sample (small 20n

ndfns

xt

Practice Midterm

.1valuep .05

thus2.093, tand 729.1between t is 1.769

),769.1P(t2value- sided- two:value-

.025.050

pp

.1valuep .05

thus2.093, tand 729.1between t is 1.769

),769.1P(t2value- sided- two:value-

.025.050

pp

Decision: since the p-value is larger than = .01, H0 is not rejected.

Conclusion: there is insufficient evidence to indicate that the true average initial checkup mileage differs from the manufacture’s recommended value.

Decision: since the p-value is larger than = .01, H0 is not rejected.

Conclusion: there is insufficient evidence to indicate that the true average initial checkup mileage differs from the manufacture’s recommended value.

Practice Midterm

(ii) 3000:H ,3000:H a0 3000:H ,3000:H a0

80.250/273

30003108

/

3000*

273s ,3108x sample), (large 50n

ns

xz 80.2

50/273

30003108

/

3000*

273s ,3108x sample), (large 50n

ns

xz

0052.)4974.5(.22.80)P(z2value-

sided- two:value-

p

p0052.)4974.5(.22.80)P(z2value-

sided- two:value-

p

p

Decision: since the p-value is smaller than = .01, H0 is rejected.

Conclusion: there is strong evidence to indicate that the true average initial checkup mileage differs from the manufacture’s recommended value.

Decision: since the p-value is smaller than = .01, H0 is rejected.

Conclusion: there is strong evidence to indicate that the true average initial checkup mileage differs from the manufacture’s recommended value.

Practice Midterm

(iii) the smallest significance level to reject the null

hypothesis=p-value=.0052.

(iv) For (i), we need to assume that the sample has been randomly selected from a normally distributed population. For (ii), the normality assumption is not needed.

Practice Midterm

6. In planning for a meeting with accounting majors, the head of the Accounting Program wants to emphasize the importance of doing well in the major courses to get better-paying jobs after graduation. To support this point, he plans to show that there is a strong relationship between starting salaries for recent accounting graduates and their grade-point average (GPA) in the major courses. Records for seven of last year's accounting graduates are selected at random:

Practice Midterm

GPA in major courses Starting salary (in thousands dollars)

2.58 16.5

3.27 18.8

3.85 19.5

3.50 19.2

3.33 18.5

2.89 16.6

2.23 15.6

5.01S 14.31,S 1.88,S 390.69,yx

2235.75,y 124.70,y 68.84,x 21.65,x

xyyyxxii

2ii

2ii

5.01S 14.31,S 1.88,S 390.69,yx

2235.75,y 124.70,y 68.84,x 21.65,x

xyyyxxii

2ii

2ii

Practice Midterm

(i) What are dependent and independent variables?(ii) Find and report the least-square regression line.(iii) How much of the variability in starting salary isexplained by the GPA in major courses?(iv) Find 95% confidence interval for the slope.Interpret the point and interval estimates of the slope.(v) Obtain a 95% confidence interval for the expectedstarting salary of all graduates with major GPA 3.0.(vi) Obtain a 95% confidence interval for a graduatewith major GPA 3.0.(vii) Suppose 5 graduates each has major GPA 3.0. Do

you expect these 5 graduates to have exactly the same starting salary?

Practice Midterm

Solution:

(i) x=Independent variable=GPA in major courses

y=dependent variable=starting salary

(ii)

x2.669.59y

9.597

21.652.66

7

124.70x β -y α

2.661.88

5.01

S

S β

xx

xy

x2.669.59y

9.597

21.652.66

7

124.70x β -y α

2.661.88

5.01

S

S β

xx

xy

Practice Midterm

(iii)

coursesmajor in GPA by the explained

issalary startingin ty variabili theof %93

93.31.1488.1

01.5

SS

S

Total SS

SSRr

S

SSSR ,S Total SS

2

yyxx

2xy2

xx

2xy

yy

coursesmajor in GPA by the explained

issalary startingin ty variabili theof %93

93.31.1488.1

01.5

SS

S

Total SS

SSRr

S

SSSR ,S Total SS

2

yyxx

2xy2

xx

2xy

yy

Practice Midterm

(iv)

When GPA increases 1 unit, the starting salary increases 2660$, and we are 95% confident that the true increase in starting salary associated with one unit GPA is between 1840$ and 3480$.

3.48) (1.84,

.82 2.66 1.88.44/ 2.57 2.66 /ˆ ˆ

is for interval confidence 95%A

44..19 σ

.19.96/52)SSE/(nσ

96.88.1/5.01-14.31 SSR -Total SSSSE

2/

2

2

xxSt

3.48) (1.84,

.82 2.66 1.88.44/ 2.57 2.66 /ˆ ˆ

is for interval confidence 95%A

44..19 σ

.19.96/52)SSE/(nσ

96.88.1/5.01-14.31 SSR -Total SSSSE

2/

2

2

xxSt

Practice Midterm

(v)

18.00). (17.14,or

.4317.571.88

21.65/7)(3

7

1.442.5717.57

S

)x(x

n

1σty

is 3at xy of average for the interval confidence 95%

17.5732.669.59y 3,When x

2

xx

20

α/2

0

0

18.00). (17.14,or

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21.65/7)(3

7

1.442.5717.57

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)x(x

n

1σty

is 3at xy of average for the interval confidence 95%

17.5732.669.59y 3,When x

2

xx

20

α/2

0

0

Practice Midterm

(vi)

(vii) No.

18.78). (16.36,or

21.157.1788.1

)7/65.213(

7

1144.57.257.17

)(11ˆˆ

is 3at xy of prediction for the interval confidence 95%

2

20

2/

0

xxS

xx

nty

18.78). (16.36,or

21.157.1788.1

)7/65.213(

7

1144.57.257.17

)(11ˆˆ

is 3at xy of prediction for the interval confidence 95%

2

20

2/

0

xxS

xx

nty