Mont4e Sm Ch09 Sec03

9-16

interval is equivalent to a two-sided hypothesis test at α = 0.01, the conclusions necessarily must be consistent.

9-43 a) 1) The parameter of interest is the true average battery life, μ. 2) H0 : μ = 4 3) H1 : μ > 4 4) α = 0.05

5) z xn0 =

− μ

σ /

6) Reject H0 if z0 > zα where z0.05 = 1.65 7) 05.4=x , σ = 0.2

77.150/2.0405.4

0 =−

=z

8) Since 1.77>1.65, reject the null hypothesis and conclude that there is sufficient evidence to conclude that the true average battery life exceeds 4 hours at α = 0.05.

b) p-value=1- )( 0ZΦ =1- )77.1(Φ ≅ 0.04

c) ⎟⎟⎠

⎞⎜⎜⎝

⎛ −−Φ=

2.050)45.4(

05.0zβ = Φ(1.65 – 17.68) = Φ(-16.03) = 0

Power = 1-β = 1-0 = 1

d) n =( ) ( )

,38.1)5.0(

)2.0()29.165.1()45.4( 2

22

2

221.005.0

2

22

=+

=−

+=

+ σδ

σβα zzzz

n ≅ 2

e) μσ≤⎟

⎠

⎞⎜⎝

⎛−n

zx 05.0

μ

μ

≤

≤⎟⎠

⎞⎜⎝

⎛−

003.4502.065.105.4

Since the lower limit of the CI is just slightly above 4, we conclude that average life is greater than 4 hours at α=0.05.

Section 9-3 9-44 a) α=0.01, n=20, the critical values are 861.2±

b) α=0.05, n=12, the critical values are 201.2± c) α=0.1, n=15, the critical values are 761.1±

9-45 a) α=0.01, n=20, the critical value = 2.539 b) α=0.05, n=12, the critical value = 1.796 c) α=0.1, n=15, the critical value = 1.345

9-46 a) α=0.01, n=20, the critical value = -2.539 b) α=0.05, n=12, the critical value = -1.796 c) α=0.1, n=15, the critical value = -1.345

9-17

9-47 a) 05.0*2025.0*2 ≤≤ p then 1.005.0 ≤≤ p

b) 05.0*2025.0*2 ≤≤ p then 1.005.0 ≤≤ p c) 4.0*225.0*2 ≤≤ p then 8.05.0 ≤≤ p

9-48 a) 05.0025.0 ≤≤ p b) 025.0105.01 −≤≤− p then 975.095.0 ≤≤ p

c) 4.025.0 ≤≤ p 9-49 a) 025.0105.01 −≤≤− p then 975.095.0 ≤≤ p

b) 05.0025.0 ≤≤ p c) 25.014.01 −≤≤− p then 75.06.0 ≤≤ p 9-50 a. 1) The parameter of interest is the true mean interior temperature life, μ. 2) H0 : μ = 22.5 3) H1 : μ ≠ 22.5 4) α = 0.05

5)ns

xt/0

μ−=

6) Reject H0 if |t0| > tα/2,n-1 where tα/2,n-1 = 2.776 7) 22.496=x , s = 0.378 n=5

00237.05/378.0

5.22496.220 −=

−=t

8) Since –0.00237 >- 2.776, we cannot reject the null hypothesis. There is not sufficient evidence to conclude that the true mean interior temperature is not equal to 22.5 °C at α = 0.05.

2*0.4 <P-value < 2* 0.5 ; 0.8 < P-value <1.0 b.) The points on the normal probability plot fall along the line. Therefore, there is no evidence to conclude that the interior temperature data is not normally distributed.

21.5 22.5 23.5

1

5

10

20304050607080

90

95

99

Data

Per

cent

Normal Probability Plot for tempML Estimates - 95% CI

Mean

StDev

22.496

0.338384

ML Estimates

9-18

c.) d = 66.0378.0

|5.2275.22||| 0 =−

=−

=σ

μμσδ

Using the OC curve, Chart VII e) for α = 0.05, d = 0.66, and n = 5, we get β ≅ 0.8 and

power of 1−0.8 = 0.2.

d) d = 66.0378.0

|5.2275.22||| 0 =−

=−

=σ

μμσδ

Using the OC curve, Chart VII e) for α = 0.05, d = 0.66, and β ≅ 0.1 (Power=0.9), 40=n . e) 95% two sided confidence interval

⎟⎟⎠

⎞⎜⎜⎝

⎛+≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛−

nstx

nstx 4,025.04,025.0 μ

965.22027.225

378.0776.2496.225

378.0776.2496.22

≤≤

⎟⎟⎠

⎞⎜⎜⎝

⎛+≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛−

μ

μ

We cannot conclude that the mean interior temperature is not equal to 22.5 since the value is included inside the confidence interval. 9-51 a. 1) The parameter of interest is the true mean female body temperature, μ. 2) H0 : μ = 98.6 3) H1 : μ ≠ 98.6 4) α = 0.05

5)ns

xt/0

μ−=


48.325/4821.0

6.98264.980 −=

−=t

8) Since 3.48 > 2.064, reject the null hypothesis and conclude that the there is sufficient evidence to

conclude that the true mean female body temperature is not equal to 98.6 °F at α = 0.05. P-value = 2* 0.001 = 0.002

b)

9-19

Data appear to be normally distributed.

c) d = 24.14821.0

|6.9898||| 0 =−

=−

=σ

μμσδ

Using the OC curve, Chart VII e) for α = 0.05, d = 1.24, and n = 25, we get β ≅ 0 and

power of 1−0 ≅ 1.

d) d = 83.04821.0

|6.982.98||| 0 =−

=−

=σ

μμσδ


⎟⎠

⎞⎜⎝

⎛+≤≤⎟⎠

⎞⎜⎝

⎛−nstx

nstx 24,025.024,025.0 μ

463.98065.9825

4821.0064.2264.9825

4821.0064.2264.98

≤≤

⎟⎠

⎞⎜⎝

⎛+≤≤⎟⎠

⎞⎜⎝

⎛−

μ

μ

We can conclude that the mean female body temperature is not equal to 98.6 since the value is not included inside the confidence interval. 9-52 a) 1) The parameter of interest is the true mean rainfall, μ. 2) H0 : μ = 25 3) H1 : μ > 25 4) α = 0.01

5) t0 = ns

x/

μ−

6) Reject H0 if t0 > tα,n-1 where t0.01,19 = 2.539 7) x = 26.04 s = 4.78 n = 20

t0 = 97.020/78.42504.26

=−

8) Since 0.97 < 2.539, do not reject the null hypothesis and conclude there is insufficient evidence to indicate that the true mean rainfall is greater than 25 acre-feet at α = 0.01. The 0.10 < P-value < 0.25.

b) the data on the normal probability plot fall along the straight line. Therefore there is evidence that the data are normally distributed.

97 98 99

1

5

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20304050607080

90

95

99

Data

Per

cent

Norm al P robab ility P lot for 9 -31M L Estimates - 95% C I

9-20

c) d = 42.078.4

|2527||| 0 =−

=−

=σ

μμσδ

Using the OC curve, Chart VII h) for α = 0.01, d = 0.42, and n = 20, we get β ≅ 0.7 and

power of 1−0.7 = 0.3.

d) d = 52.078.4

|255.27||| 0 =−

=−

=σ

μμσδ

Using the OC curve, Chart VII h) for α = 0.05, d = 0.42, and β ≅ 0.1 (Power=0.9), 75=n . e) 99% lower confidence bound on the mean diameter

0.01,19sx tn

μ⎛ ⎞− ≤⎜ ⎟⎝ ⎠

4.7826.04 2.53920

23.326

μ

μ

⎛ ⎞− ≤⎜ ⎟⎝ ⎠

≤

Since the lower limit of the CI is less than 25, we conclude that there is insufficient evidence to indicate that the true mean rainfall is greater than 25 acre-feet at α = 0.01.

9-53 a)

1) The parameter of interest is the true mean sodium content, μ. 2) H0 : μ = 130 3) H1 : μ ≠ 130 4) α = 0.05

5)ns

xt/0

μ−=

403020

99

95

90

80706050403020

10

5

1

Data

Per

cent

Normal Probability Plot for rainfallML Estimates - 95% CI

Mean

StDev

26.035

4.66361

ML Estimates

9-21


456.130/929.0130753.129

0 −=−

=t

8) Since 1.456 < 2.064, do not reject the null hypothesis and conclude that the there is not sufficient evidence that the true mean sodium content is different from 130mg at α = 0.05.

From table V the t0 value is found between the values of 0.05 and 0.1 with 29 degrees of freedom, so 2*0.05<P-value < 2* 0.1 Therefore, 0.1< P-value < 0.2.

b) The assumption of normality appears to be reasonable.

c) d = 538.0929.0

|1305.130||| 0 =−

=−

=σ

μμσδ

Using the OC curve, Chart VII e) for α = 0.05, d = 0.53, and n = 30, we get β ≅ 0.2 and power

of 1−0.20 = 0.80

d) d = 11.0929.0

|1301.130||| 0 =−

=−

=σ

μμσδ


⎟⎟⎠

⎞⎜⎜⎝

⎛+≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛−

nstx

nstx 29,025.029,025.0 μ

100.130406.12930929.0045.2753.129

30929.0045.2753.129

≤≤

⎟⎟⎠

⎞⎜⎜⎝

⎛+≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛−

μ

μ

132131130129128127

99

95

90

80706050403020

10

5

1

Data

Per

cent

Normal Probability Plot for 9-33ML Estimates - 95% CI

9-22

There is no evidence that the mean differs from 130 because that value is inside the confidence interval.

9-54 a) In order to use t statistics in hypothesis testing, we need to assume that the underlying distribution is normal.

1) The parameter of interest is the true mean coefficient of restitution, μ. 2) H0 : μ = 0.635 3) H1 : μ > 0.635 4) α = 0.05

5) t0 = ns

x/

μ−


t0 = 35.540/013.0635.0624.0

−=−

8) Since –5.25 < 1.685, do not reject the null hypothesis and conclude that there is not sufficient evidence to indicate that the true mean coefficient of restitution is greater than 0.635 at α = 0.05. The area to right of -5.35 under the t distribution is greater than 0.9995 from table V.

Minitab gives P-value = 1. b) From the normal probability plot, the normality assumption seems reasonable:

Baseball Coeff of Restitution

Perc

ent

0.660.650.640.630.620.610.600.59

99

95

90

80

70

60504030

20

10

5

1

Probability Plot of Baseball Coeff of RestitutionNormal

c) d = 38.0013.0

|635.064.0||| 0 =−

=−

=σ

μμσδ

Using the OC curve, Chart VII g) for α = 0.05, d = 0.38, and n = 40, we get β ≅ 0.25 and power of 1−0.25 = 0.75.

d) d = 23.0013.0

|635.0638.0||| 0 =−

=−

=σ

μμσδ

Using the OC curve, Chart VII g) for α = 0.05, d = 0.23, and β ≅ 0.25 (Power=0.75), 40=n .

9-23

e) Lower confidence bound is 6205.01, =⎟⎠

⎞⎜⎝

⎛− − nstx nα

Since 0.635 > 0.6205, then we fail to reject the null hypothesis.

9-55 a) In order to use t statistics in hypothesis testing, we need to assume that the underlying distribution is normal.

1) The parameter of interest is the true mean oxygen concentration, μ. 2) H0 : μ = 4 3) H1 : μ ≠ 4 4) α = 0.01

5) t0 = ns

x/

μ−

6) Reject H0 if |t0 |>tα/2, n-1 = t0.005, 19 = 2.861 7) ⎯x = 3.265, s = 2.127, n = 20

t0 = 55.120/127.24265.3

−=−

8) Because -2.861<-1.55 do not reject the null hypothesis and conclude that there is insufficient evidence to indicate that the true mean oxygen differs from 4 at α = 0.01. P-Value: 2*0.05<P-value<2*0.10 therefore 0.10< P-value<0.20

b) From the normal probability plot, the normality assumption seems reasonable:

O2 concentration

Perc

ent

7.55.02.50.0

99

95

90

80

70

60504030

20

10

5

1

Probability Plot of O2 concentrationNormal

c.) d = 47.0127.2

|43||| 0 =−

=−

=σ

μμσδ

Using the OC curve, Chart VII f) for α = 0.01, d = 0.47, and n = 20, we get β ≅ 0.70 and power of 1−0.70 = 0.30.

d) d = 71.0127.2

|45.2||| 0 =−

=−

=σ

μμσδ

Using the OC curve, Chart VII f) for α = 0.01, d = 0.71, and β ≅ 0.10 (Power=0.90), 40=n .

e) The 95% confidence interval is:

9-24

⎟⎠

⎞⎜⎝

⎛+≤≤⎟⎠

⎞⎜⎝

⎛− −− nstx

nstx nn 1,2/1,2/ αα μ = 62.49.1 ≤≤ μ

Because 4 is within the confidence interval, we fail to reject the null hypothesis.

9-56 a) 1) The parameter of interest is the true mean sodium content, μ.

2) H0 : μ = 300 3) H1 : μ > 300 4) α = 0.05

5)ns

xt/0

μ−=

6) Reject H0 if t0 > tα,n-1 where tα,n-1 = 1.943 7) 315=x , s = 16 n=7

48.27/16

3003150 =

−=t

8) Since 2.48>1.943, reject the null hypothesis and conclude that there is sufficient evidence that the leg strength exceeds 300 watts at α = 0.05. The p-value is between .01 and .025

b) d = 3125.016

|300305||| 0 =−

=−

=σ

μμσδ

Using the OC curve, Chart VII g) for α = 0.05, d = 0.3125, and n = 7, β ≅ 0.9 and power = 1−0.9 = 0.1.

c) if 1-β>0.9 then β<0.1 and n is approximately 100

d) Lower confidence bound is 2.3031, =⎟⎠

⎞⎜⎝

⎛− − nstx nα

because 300< 303.2 reject the null hypothesis

9-57 a.)1) The parameter of interest is the true mean tire life, μ. 2) H0 : μ = 60000 3) H1 : μ > 60000 4) α = 0.05

5) t0 = ns

x/

μ−

6) Reject H0 if t0 > tα,n-1 where 753.115,05.0 =t

7) 94.36457.139,6016 === sxn

t0 = 15.016/94.3645

600007.60139=

−

8) Since 0.15 < 1.753., do not reject the null hypothesis and conclude there is insufficient evidence to indicate that the true mean tire life is greater than 60,000 kilometers at α = 0.05. The P-value > 0.40.

b.) d = 27.094.3645

|6000061000||| 0 =−

=−

=σ

μμσδ

Using the OC curve, Chart VII g) for α = 0.05, d = 0.27, and β ≅ 0.1 (Power=0.9),

9-25

4=n . Yes, the sample size of 16 was sufficient.

9-58 In order to use t statistics in hypothesis testing, we need to assume that the underlying distribution is normal. 1) The parameter of interest is the true mean impact strength, μ. 2) H0 : μ = 1.0 3) H1 : μ > 1.0 4) α = 0.05

5) t0 = ns

x/

μ−


t0 = 47.420/25.00.125.1

=−

8) Since 4.47 > 1.729, reject the null hypothesis and conclude there is sufficient evidence to indicate that the true mean impact strength is greater than 1.0 ft-lb/in at α = 0.05. The P-value < 0.0005

9-59 In order to use t statistic in hypothesis testing, we need to assume that the underlying distribution is normal. 1) The parameter of interest is the true mean current, μ. 2) H0 : μ = 300 3) H1 : μ > 300 4) α = 0.05

5) t0 = ns

x/

μ−

6) Reject H0 if t0 > tα,n-1 where 833.19,05.0 =t

7) 7.152.31710 === sxn

t0 = 46.310/7.153002.317

=−

8) Since 3.46 > 1.833, reject the null hypothesis and conclude there is sufficient evidence to indicate that the true mean current is greater than 300 microamps at α = 0.05. The 0.0025 <P-value < 0.005 9-60 a)

1) The parameter of interest is the true mean height of female engineering students, μ. 2) H0 : μ = 65 3) H1 : μ > 65 4) α = 0.05

5) t0 = ns

x/

μ−

6) Reject H0 if t0 > tα,n-1 where t0.05,36 =1.68 7) 65.811 =x inches 2.106=s inches n = 37

t0 = 34.237/11.265811.65

=−

8) Since 2.34 > 1.68, reject the null hypothesis and conclude that there is sufficient evidence to indicate that the true mean height of female engineering students is not equal to 65 at α = 0.05. P-value: 0.01<P-value<0.025.

b.) From the normal probability plot, the normality assumption seems reasonable:

9-26

Female heights

Perc

ent

72706866646260

99

95

90

80

70

60504030

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10

5

1

Probability Plot of Female heightsNormal

c) 42.111.2

6562=

−=d , n=37 so, from the OC Chart VII g) for α = 0.05, we find that β≅0.

Therefore, the power ≅ 1.

d.) 47.011.2

6564=

−=d so, from the OC Chart VII g) for α = 0.05, and β≅0.2 (Power=0.8).

30* =n . 9-61 a) In order to use t statistics in hypothesis testing, we need to assume that the underlying distribution is normal.

1) The parameter of interest is the true mean distance, μ. 2) H0 : μ = 280 3) H1 : μ > 280 4) α = 0.05

5) t0 = ns

x/

μ−

6) Reject H0 if t0 > tα,n-1 where t0.05,99 =1.6604 7) x = 260.3 s = 13.41 n = 100

t0 = 69.14100/41.132803.260

−=−

8) Since –14.69 < 1.6604, do not reject the null hypothesis and conclude that there is insufficient evidence to indicate that the true mean distance is greater than 280 at α = 0.05. From table V the t0 value in absolute value is greater than the value corresponding to 0.0005. Therefore, the P-value is greater than 0.9995.


9-27

Distance for golf balls

Perc

ent

310300290280270260250240230220

99.9

99

9590

80706050403020

10

5

1

0.1

Probability Plot of Distance for golf ballsNormal

c) d = 75.041.13

|280290||| 0 =−

=−

=σ

μμσδ

Using the OC curve, Chart VII g) for α = 0.05, d = 0.75, and n = 100, β ≅ 0 and power of 1−0 = 1.

d) d = 75.041.13

|280290||| 0 =−

=−

=σ

μμσδ

Using the OC curve, Chart VII g) for α = 0.05, d = 0.75, and β ≅ 0.20 (Power=0.80), 15=n . 9-62 a) In order to use t statistics in hypothesis testing, we need to assume that the underlying distribution is normal. 1) The parameter of interest is the true mean concentration of suspended solids, μ. 2) H0 : μ = 55 3) H1 : μ ≠ 55 4) α = 0.05

5) t0 = ns

x/

μ−

6) Reject H0 if |t0 | > tα/2,n-1 where t0.025,59 =2.000 7) x = 59.87 s = 12.50 n = 60

t0 = 018.360/50.125587.59

=−

8) Since 3.018 > 2.000, reject the null hypothesis and conclude that there is sufficient evidence to indicate that the true mean concentration of suspended solids is not equal to 55 at α = 0.05.

From table V the t0 value is between the values of 0.001 and 0.0025 with 59 degrees of freedom. Therefore 2*0.001<P-value < 2* 0.0025 and 0.002< P-value<0.005. Minitab gives a P-value of 0.0038.


9-28

Concentration of solids

Perc

ent

1009080706050403020

99.9

99

9590

80706050403020

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5

1

0.1

Probability Plot of Concentration of solidsNormal

d) 4.050.125550

=−

=d , n=60 so, from the OC Chart VII e) for α = 0.05, d= 0.4 and n=60 we

find that β≅0.2. Therefore, the power = 1-0.2 = 0.8. e) From the same OC chart, and for the specified power, we would need approximately 75 observations.

4.050.125550

=−

=d Using the OC Chart VII e) for α = 0.05, d = 0.4, and β ≅ 0.10 so

that power=0.90, 75=n .

.

Documents

Mont4e Sm Ch09 Sec03