Stats151 Spring 2012 Assignment 3 Solutions All problems are ...

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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 ...
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 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 8.8

(a)[1 mark]

_ n = 45; x = $129,849/45 = $2885.5

(b)[4 marks]

The confidence interval will be

x

2 / n to x

2 / n

$2885 .5 2($1350 ) / 45 to $2885 .5 2($1350 ) / 45 $2483 .0 to $3288 .0 (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

x

z

/2

/ n

to x

z

/2

/ n

947.4 1.96(188) / 18 to 947.4 1.96(188) / 18 860.5 to 1034.3 mg per day (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; z

/2

= z0.005 = 2.575

Step 2:

x

z

/ n to x

/2

2.27 2.575(0.5) / 30

z

/2

/ n

to 2.27 2.575(0.5) / 30

2.03 to 2.51 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. 2

z

n

1.96(0.) 5 0.1

/2

E

(d)[3 mark]

2

96.04

97

(in $millions)

x

z

/ n to x

/2

2.35 1.96(0.5) / 97

z

/2

/ n

to 2.35 1.96(0.5) / 97

2.25 to 2.45 8.98 n = 38; df = 37; t

/2

_ = t0.05 = 1.688; x = 5.6 pmol/l; s = 1.9

pmol/ (a)[4 marks]

x

t

/2

s / n to x

5.6 1.688(1.9) / 38

t

/2

s/ n

to 5.6 1.688(1.9) / 38

5.08 to 6.12

(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 (a)

H0:

denote the mean post-work heart rate of casting workers. = 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 Step 1:

H0:

Step 2: Step 3:

= 78.3,

= 11.2

= 72 bpm, Ha: = 0.05

z

Step 4:

(78.3 72) /(11.2 / 29)

3.03

Critical-Value Approach: P-Value Approach:

Step 5:

Critical value =

P-value is P( Z

Critical-Value Approach: P-value Approach:

Step 6:

> 72 bpm

z

z0.05 1.645 .

3.03) 1 0.9988 0.0012 .

Since 3.03 > 1.645, reject H0.

Since 0.0012 < 0.05, reject H0.

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:

Step 2:

= $1874, Ha:

$1874

= 0.05

2060.76 1874 350.9 / 25

Step 3:

t

2.661

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
0.10 Step 5: Since –0.886 > -1.660, do not reject H0. Since the P-value > Step 6:

, do not reject H0.

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.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 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

spent by teens and

= 2

2

, Ha:

1


1.671, reject H0.

(b)[3 marks] 90% CI = (20 18)

1.671(4.6002) (1/ 30 1/ 40)

(0.143,3.857)

Control, n1 = 74, x1 = 84.4, s1 = 12.6

10.44 [6 marks] Population 1:

Dexamethasone, n2 = 72, x2 = 78.2, s2 = 15.0

Population 2:

sp

73(12.6)2 71(15.0)2 144

Step 1:

H0:

Step 2: Step 3:

Step 4:

2,

1

191.42 Ha:

13.8355

1

2

= 0.01

84.4 78.2

t

2.707

13.8355 1/ 74 1/ 72

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, x1 = 67.9, s1 = 5.49 - = 66.81, s = 9.04 Population 2: Control, n2 = 31, x 2 2 (a)[6 marks]

H0:

Step 2: Step 3:

Step 4:

2,

1

Ha:

1

2

= 0.05

t

67.90 66.81

0.459

(5.492 /10) (9.042 / 31)

s12 n1 2 1

s n1

s22 n2 2

n1 1

2 2

s n2

2

2

n2 1

5.492 10 5.49 10

10 1

Critical value = -1.708 P-value > 0.100.

2

2

9.042 31 2

9.04 31

2

31 1

2

25.72; df

25

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;

[2 marks]Step 1: Step 2:

H0:

= 0.01

Population 2: 1

2,

Ha:

Before; 1

2

df = 13

[6 marks]Step 3:

d

(

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

d)/ n

797 .9 / 14

d sd / n

t [1 mark]Step 4:

56 .99 , sd

56 .99 26 .20 / 14

26 .20

8.14

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.

56.99 2.650

26.20 148

56.99 18.56

(38.43, 75.56) nmo/L

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