Stats151 Spring 2012 Assignment 3 Solutions

All problems are taken from the textbook, Introductory Statistics by Neil A Weiss, 9th

Edition. Solve the following problems:

8.8, 8.32, 8.36, 8.66, 8.98

9.10, 9.58, 9.78, 9.104, 9.106

10.10, 10.38, 10.44, 10.72, 10.146, 10.152

Total Marks=98

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8.8 (a)[1 mark] n = 45; x_ = $129,849/45 = $2885.5

(b)[4 marks] The confidence interval will be

0.3288$0.2483$

45/)1350($25.2885$45/)1350($25.2885$

/2/2

to

to

nxtonx

(c)[1 mark] We could see if a histogram looked bell-shaped or

if a normal probability plot produced a relatively straight

line.

(d)[2 marks] It is not necessary that the budgets for home

improvement costs be exactly normally distributed since the

sample size 45 is large enough to ensure that the interval

obtained is approximately correct.

8.32 (a)[4 marks] The sample mean is 17053 / 18 = 947.4 mg per day.

The 95% confidence interval for µ is

/ 2 / 2/ /

947.4 1.96(188) / 18 947.4 1.96(188) / 18

860.5 1034.3 mg per day

x z n to x z n

to

to

(b)[2 marks] We can be 95% confident that the interval from

860.5 to 1034.3 mg per day contains the population mean daily

calcium intake for adults.

8.36 [5 marks] n = 30; x_ = $2.27 million; = $0.5 million

Step 1: = 0.01; / 2z = z0.005 = 2.575

Step 2:

/ 2 / 2/ to /

2.27 2.575(0.5) / 30 to 2.27 2.575(0.5) / 30

2.03 to 2.51

x z n x z n

We can be 99% confident that the mean gross earnings, , of all

Rolling Stones concerts is somewhere between $2.03 and $2.51

million.

8.66 (a)[1 mark] E = (2.51 - 2.03)/2 = 0.24 million

(b)[1 mark] We can be 99% confident that the maximum error made

in using x_ to estimate is $0.24 million.

(c)[2 marks] The margin of error of the estimate is specified to

be E = $0.1 million.

nz

E

/ . (.)

..

2

2 21 96 0 5

0 196 04 97

(d)[3 mark] (in $millions)

/ 2 / 2/ to /

2.35 1.96(0.5) / 97 to 2.35 1.96(0.5) / 97

2.25 to 2.45

x z n x z n

8.98 n = 38; df = 37; / 2t = t0.05 = 1.688; x_ = 5.6 pmol/l; s = 1.9

pmol/

(a)[4 marks]

/ 2 / 2/ to /

5.6 1.688(1.9) / 38 to 5.6 1.688(1.9) / 38

5.08 to 6.12

x t s n x t s n

(b)[2 marks] We can be 90% confident that the mean plasma level

of adrenomedullin, , for women with recurrent pregnancy

loss is somewhere between 5.08 and 6.12 pmol/l.

9.10 [3 marks] Let denote the mean post-work heart rate of casting workers.

(a) H0: = 72 beats/min (b) Ha: > 72 beats/min (c) right-tailed

9.58 (a)[1 mark] z = 3.08, Right-tail probability = 1.0000 - 0.9990 = 0.0010

P-value = 0.001 x 2 = 0.0020

At a 5% significance level, we would reject the null hypothesis in favor of the alternative

hypothesis.

(b)[1 mark] z = -2.42, Left-tail probability = 0.0078

P-value = 0.0078 x 2 = 0.0156

At a 5% significance level, we would reject the null hypothesis in favor of the alternative

hypothesis.

9.78 [6 marks] n = 29, x_ = 78.3, = 11.2

Step 1: H0: = 72 bpm, Ha: > 72 bpm

Step 2: = 0.05

Step 3: (78.3 72) /(11.2 / 29) 3.03z

Step 4: Critical-Value Approach: Critical value = 0.05 1.645z z .

P-Value Approach: P-value is ( 3.03) 1 0.9988 0.0012P Z .

Step 5: Critical-Value Approach: Since 3.03 > 1.645, reject H0.

P-value Approach: Since 0.0012 < 0.05, reject H0.

Step 6: At the 5% significance level, the data provide sufficient

evidence to conclude that the mean post-work heart rate for

casting workings exceeds the normal resting heart rate of 72

beats per minute.

9.104 [6 marks] n = 25, df = 24, x_ = $2060.76, s = $350.90

Step 1: H0: = $1874, Ha: $1874

Step 2: = 0.05

Step 3: 2060.76 1874

2.661350.9 / 25

t

Step 4: Critical Value Approach: Critical values = +2.064

P-Value Approach: 0.01 < P-value < 0.02

Step 5: Since 2.661 > 2.064, reject H0. Since the P-value < , reject H0.

Step 6: At the 5% significance level, the data do provide sufficient evidence to conclude that the

mean annual expenditure on apparel and services for consumer units in the Northeast

differed from the national mean in 2006.

9.106 [6 marks] n = 200, df = 199, x_ = $2480, s = $766

Step 1: H0: = $2528, Ha: < $2528

Step 2: = 0.05

Step 3: 2480 2528

0.886766 / 200

t

Step 4: Critical Value Approach: Critical value = -1.660

P-Value Approach: P > 0.10

Step 5: Since –0.886 > -1.660, do not reject H0. Since the P-value > , do not reject H0.

Step 6: At the 5% significance level, the data do not provide

sufficient evidence to conclude that the mean cost of

having a baby by AML is less than the average cost of

having a baby in a U.S. hospital.

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10.10 (a)[1 mark] The variable is the amount spent at shopping malls

(b)[1 mark] The two populations are teens and adults.

(c)[2 marks] H0: 1 = 2 , Ha: 1 < 2 where 1 is the mean amount

spent by teens and 2 is the mean amount spent by adults.

(d)[1 mark] left-tailed

10.38 (a)[5 mark]

2 229(4) 39(5)21.16176 4.6002

30 40 2ps

Step 1: H0: 1 2, Ha: 1 2

Step 2: = 0.05

Step 3: 20 18

1.8004.6002 (1/ 30) (1/ 40)

t

Step 4: df = 68, Critical value = 1.668

0.025 < P-value < 0.050.

Step 5: Since 1.800 > 1.671, reject H0.

(b)[3 marks] 90% CI = (20 18) 1.671(4.6002) (1/ 30 1/ 40) (0.143,3.857)

10.44 [6 marks] Population 1: Control, n1 = 74, x- 1 = 84.4, s1 = 12.6

Population 2: Dexamethasone, n2 = 72, x- 2 = 78.2, s2 = 15.0

2 273(12.6) 71(15.0)191.42 13.8355

144ps

Step 1: H0: 1 2, Ha: 1 2

Step 2: = 0.01

Step 3: 84.4 78.2

2.70713.8355 1/ 74 1/ 72

t

Step 4: df = 144, Critical value = 2.353 (using technology)

For the P-value approach, P(t > 2.707) < 0.005.

Step 5: Since 2.707 > 2.353, reject H0. Because the P-value

is less than the significance level, reject H0.

Step 6: At the 1% significance level, the data provide

sufficient evidence that early dexamethasone therapy

has, on average, an adverse effect on IQ.

10.72 Population 1: Intervention, n1 = 10, x- 1 = 67.9, s1 = 5.49

Population 2: Control, n2 = 31, xx- 2 = 66.81, s2 = 9.04

(a)[6 marks] H0: 1 2, Ha: 1 2

Step 2: = 0.05

Step 3: 2 2

67.90 66.810.459

(5.49 /10) (9.04 / 31)t

Step 4:

2 22 2 2 21 2

1 2

2 2 2 22 2 2 2

1 2

1 2

1 2

5.49 9.04

10 3125.72; 25

5.49 9.04

10 31

10 1 31 11 1

s s

n ndf

s s

n n

n n

Critical value = -1.708

P-value > 0.100.

Step 5: Since 0.459 > -1.708, do not reject H0.

At the 5% significance level, the data do not provide sufficient evidence to conclude that

the intervention program reduces mean heart rate of urban bus drivers in Stockholm.

For the P-value approach, 2P(t < 0.459) > 0.10. Therefore, since

the P-value is larger than the significance level, do not reject

H0.

(b)[2 marks]One possibility is that the new routes were mistakenly

judged to be ‘improved.’ There are several other possibilities,

depending on when the data were collected. If the data were

collected soon after the new assignments, it may be that the

intervention routes, even if improved, were still not as

comfortable for the drivers as were the normal routes for the

drivers still on those routes. Heart rate is also affected by

other factors, such as the driver’s weight and overall condition of

health, time elapsed since the last meal, amount of caffeine

consumed, and having to deal with new and unfamiliar passengers.

We can’t tell whether there was any attempt to control for these

factors. If the drivers were chosen for the improved routes, not

at random, but because they were already judged to be under more

stress than those left on the normal routes, there would have been

an underlying health condition that may not have responded to the

route changes. Still another possibility is that heart rate is not

as closely related to stress as some other variable such as blood

pressure.

(c)[1 mark]Although we are lacking information on how the drivers were

chosen for the improved routes, it appears that this was a designed

experiment. If so, base line heart rates (and possibly other data)

should have been collected so that it would be possible to

determine whether there was a lowering of the heart rate in the

intervention group.

10.146 Population 1: After; Population 2: Before; df = 13

[2 marks]Step 1: H0: 1 2, Ha: 1 2

Step 2: = 0.01

26.2056.99 2.650 56.99 18.56 (38.43,75.56) nmo/L

148

[6 marks]Step 3: The paired differences, d = x2 – x1, are

25.2 104.7 49.9 32.3 71.1 23.4 65.8

71.1 81.8 53.4 41.1 87.1 19.8 71.2

99.5614/9.797/)( ndd , 20.26ds

14.814/20.26

99.56

/ ns

dt

d

[1 mark]Step 4: Critical value = 2.650

P-Value < 0.005

[1 mark]Step 5: Since 8.14 > 2.650, reject H0.

For the P-value approach, since the P-value is

smaller than the significance level, reject H0.

[1 mark]Step 6: At the 1% significance level, the data provide

sufficient evidence to conclude that drinking

fortified orange juice increases the serum 25(OH)D

concentration in the blood.

Note: If the populations were reversed so that Population 1

is Before and Population 2 is After, you would be

doing a left tailed test. You would have a test

statistic of -8.14, a critical value of -2.650,

but the conclusion would be the same.

10.152 [4 marks]From Exercise 10.146, , df = 13.

We can be 98% confident that the mean increase, 2 1, in serum

25(OH)D concentration in the blood after drinking Vitamin D

fortified orange juice for 12 weeks is between 38.43 and 75.56

nmo/L