Chapter 8: Estimation: Single Population
a. Check for nonnormality Probability Plot of Ex_8.1 Normal 99
Mean StDev N AD P-Value
95 90
7 2.268 8 0.185 0.865
80
Percent
8.1
70 60 50 40 30 20 10 5
1
2
4
6
8
10
12
Ex_8.1
The distribution shows no significant evidence of nonnormality. b. Point estimate of the population mean that is unbiased, efficient and consistent. ∑ X i = 560 = 7.0 Unbiased point estimator is the sample mean: X = n 8 c. The unbiased point estimate of the variance of the sample mean: s 2 (2.268) 2 Var ( X ) = = = .64298 8 n
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8.2 a. Evidence of non-normality? Probability Plot of Homes_Ex8.2 Normal 99
Mean StDev N AD P-Value
95 90
101.4 14.20 8 0.194 0.837
Percent
80 70 60 50 40 30 20 10 5
1
60
70
80
90
100 110 Homes_Ex8.2
120
130
140
No evidence of non-normality. b. The minimum variance unbiased point estimator of the population ∑ X i = 560 = 101.375 mean is the sample mean: X = n 8 c. The unbiased point estimate of the variance of the sample mean: s 2 = 201.6964 σ2 s 2 201.6964 ˆ (X ) = = Var ( X ) = ; Var = 25.2121 n n 8 x 3 d. pˆ = = = .375 n 8
8.3
n = 10 economists forecast for percentage growth in real GDP in the next year
Descriptive Statistics: RGDP_Ex8.3 Variable RGDP_Ex8.3
N 10
N* 0
Variable RGDP_Ex8.3
Minimum 2.2000
Mean 2.5700 Q1 2.4000
SE Mean 0.0716
TrMean 2.5625
StDev 0.2263
Variance 0.0512
Median 2.5500
Q3 2.7250
Maximum 3.0000
Range 0.8000
CoefVar 8.81 IQR 0.3250
a. Unbiased point estimator of the population mean is the sample mean: ∑ X i = 2.57 X= n b. The unbiased point estimate of the population variance: s 2 = .0512 c. Unbiased point estimate of the variance of the sample mean s 2 .0512 Var ( X ) = = = .00512 n 10
Sum 25.7000
Chapter 8: Estimation: Single Population
x 7 = = .70 n 10 e. Unbiased estimate of the variance of the sample proportion: pˆ (1 − pˆ ) .7(1 − .7) Var ( pˆ ) = = = .021 n 10
d. Unbiased estimate of the population proportion: pˆ =
8.4 n = 12 employees. Number of hours of overtime worked in the last month: a. Unbiased point estimator of the population mean is the sample mean: ∑ X i = 24.42 X= n b. The unbiased point estimate of the population variance: s 2 = 85.72 c. Unbiased point estimate of the variance of the sample mean s 2 85.72 Var ( X ) = = = 7.1433 n 12 x 3 d. Unbiased estimate of the population proportion: pˆ = = = .25 n 12 e. Unbiased estimate of the variance of the sample proportion: pˆ (1 − pˆ ) .25(1 − .25) Var ( pˆ ) = = = .015625 n 12 a. Check each variable for normal distribution:
Normal Probability Plot
.999 .99 .95
Probability
8.5
.80 .50 .20 .05 .01 .001 45
50
55
Meals Average: 50.1 StDev: 2.46842 N: 30
Anderson-Darling Normality Test A-Squared: 0.413 P-Value: 0.318
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Normal Probability Plot
.999 .99
Probability
.95 .80 .50 .20 .05 .01 .001 15
20
25
Attendance Average: 21.24 StDev: 2.50466 N: 25
Anderson-Darling Normality Test A-Squared: 0.377 P-Value: 0.383
Normal Probability Plot
.999 .99 .95
Probability
166
.80 .50 .20 .05 .01 .001 9
10
11
12
13
14
15
16
Ages Average: 12.24 StDev: 1.96384 N: 25
Anderson-Darling Normality Test A-Squared: 0.569 P-Value: 0.126
No evidence of non-normality in Meals, Attendance or Ages
Chapter 8: Estimation: Single Population
167
b. Unbiased estimates of population mean and variance: Descriptive Statistics: Meals, Attendance, Ages Variable Mean Meals 0.451 Attendan 0.501 Ages 0.393 Variable Meals Attendan Ages
N
Mean
Median
TrMean
StDev
30
50.100
50.000
50.192
2.468
25
21.240
21.000
21.348
2.505
25
12.240
12.000
12.217
1.964
Minimum 45.000 15.000 9.000
Maximum 55.000 25.000 16.000
Q1 48.000 19.500 10.000
Q3 52.000 23.500 14.000
Variable Unbiased estimate of mean Unbiased estimate of variance (s2) Meals 50.100 (2.468)2 = 6.0910 Attendance 21.240 (2.505)2 = 6.2750 Ages 12.240 (1.964)2 = 3.8573
8.6
1 1 μ μ E ( X1 ) + E ( X 2 ) = + = μ 2 2 2 2 1 3 μ 3μ E (Y ) = E ( X 1 ) + E ( X 2 ) = + =μ 4 4 4 4 1 2 μ 2μ E (Z ) = E ( X1 ) + E ( X 2 ) = + =μ 3 3 3 3 σ2 1 1 1σ2 σ2 b. Var ( X ) = = Var ( X 1 ) + Var ( X 2 ) = = 4 2 8 4 n 4 2 1 9 5σ Var (Y ) = Var ( X 1 ) + Var ( X 2 ) = 16 16 8 2 1 4 5σ Var ( Z ) = Var ( X 1 ) + Var ( X 2 ) = 9 9 9 X is most efficient since Var ( X ) < Var (Y ) < Var ( Z ) Var (Y ) 5 = = 2.5 c. Relative efficiency between Y and X : Var ( X ) 2 Var ( Z ) 20 Relative efficiency between Z and X : = = 2.222 Var ( X ) 9 a. E ( X ) =
SE
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8.7 a. Evidence of non-normality? Normal Probability Plot for Leak Rates ( ML Estimates
99
ML Estimates
95
Mean
0.0515
StDev
0.0216428
90
Goodness of Fit
Percent
80
AD*
70 60 50 40 30
0.596
20 10 5 1 0.00
0.05
0.10
Data
No evidence of nonnormality exists. b. The minimum variance unbiased point estimator of the population ∑ X i = .0515 mean is the sample mean: X = n c. The unbiased point estimate of the variance of the sample mean: s 2 x = (.0216428) 2 = .0004684
Var ( X ) = 8.8
σ2 n
ˆ (X ) = ; Var
s 2 .0004684 = = .00000937 n 50
a. Evidence of non-normality? No evidence of the data distribution coming from a non-normal population b. The minimum variance unbiased point estimator of the population ∑ X i = 3.8079 mean is the sample mean: X = n Descriptive Statistics: Volumes Variable Mean Volumes 0.0118 Variable Volumes
N
Mean
Median
TrMean
StDev
75
3.8079
3.7900
3.8054
0.1024
Minimum 3.5700
Maximum 4.1100
Q1 3.7400
Q3 3.8700
SE
Chapter 8: Estimation: Single Population
c. Minimum variance unbiased point estimate of the population variance is the sample variance s2 = .0105 8.9 Reliability factor for each of the following: a. 96% confidence level: zα 2 = +/- 2.05 b. 88% confidence level: zα 2 = +/- 1.56 c. 85% confidence level: zα 2 = +/- 1.44 d. α = .07 zα 2 = +/- 1.81 e. α 2 = .07 zα 2 = +/- 1.48 8.10 Calculate the margin of error to estimate the population mean a. 98% confidence level; n = 64, variance = 144 ⎞ = 3.495 ME = zα 2 σ = 2.33 ⎛⎜12 ⎟ n 64 ⎠ ⎝ b. 99% confidence interval, n=120; standard deviation = 100 ⎞ = 23.552 ME = zα 2 σ = 2.58 ⎛⎜100 ⎟ n 120 ⎠ ⎝ 8.11 Calculate the width to estimate the population mean, for a. 90% confidence level, n = 100, variance = 169 ⎞ ⎤ = 4.277 ⎤ = 2 ⎡1.645 ⎛13 width = 2ME = 2 ⎡⎢ zα 2 σ ⎜ ⎟ ⎢ ⎥ 100 ⎠ ⎥⎦ n⎦ ⎝ ⎣ ⎣ b. 95% confidence interval, n = 120, standard deviation = 25 ⎞ ⎤ = 8.9461 ⎤ = 2 ⎡1.96 ⎛ 25 width = 2ME = 2 ⎡⎢ zα 2 σ ⎜ ⎟ ⎢ ⎥ 120 ⎠ ⎥⎦ n⎦ ⎝ ⎣ ⎣ 8.12
Calculate the LCL and UCL ⎞ = 40.2 to 59.8 a. x ± zα 2 σ = 50 ± 1.96 ⎛⎜ 40 ⎟ n 64 ⎠ ⎝ ⎞ = 81.56 to 88.44 = 85 ± 2.58 ⎛⎜ 20 b. x ± zα 2 σ ⎟ n 225 ⎠ ⎝ ⎞ = 506.2652 to 513.73478 = 510 ± 1.645 ⎛⎜ 50 c. x ± zα 2 σ ⎟ n 485 ⎠ ⎝
8.13 a. n = 9, x = 187.9, σ = 32.4, z.10 = 1.28 187.9 ± 1.28(32.4/3) = 174.076 up to 201.724 b. 210.0 – 187.9 = 22.1 = zα / 2 (32.4 / 3), zα / 2 = 2.05 α = 2[1 − Fz (2.05)] = .0404 100(1-.0404)% = 95.96%
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8.14 a. Find the reliability factor for 92% confidence level: zα 2 = +/- 1.75 b. Calculate the standard error of the mean σ = 6 = .63246 90 n c. Calculate the width of the 92% confidence interval for the population mean ⎞ ⎤ = 2.2136 ⎤ = 2 ⎡1.75 ⎛ 6 width = 2ME = 2 ⎡⎢ zα 2 σ ⎜ ⎟ ⎢ ⎥ 90 ⎠ ⎥⎦ n⎦ ⎝ ⎣ ⎣ 8.15 a. n = 25, x = 2.90, σ = .45, z.025 = 1.96 ⎞ = 2.90 ± 1.96(.45/5) = 2.7236 up to 3.0764 x ± z ⎛⎜ σ ⎟ n ⎝ ⎠ b. 2.99 – 2.90 = .09 = zα / 2 (.45 / 5), zα / 2 = 1
α = 2[1 − Fz (1)] = .3174 100(1-.3174)% = 68.26% 8.16 a. n = 16, x = 4.07, σ = .12, z.005 = 2.58 4.07 ± 2.58(.12/4) = 3.9926 up to 4.1474 b. narrower since the z score for a 95% confidence interval is smaller than the z score for the 99% confidence interval c. narrower due to the smaller standard error d. wider due to the larger standard error 8.17 Find the reliability factor tv ,α 2 a. b. c. d.
n = 20, 90% confidence level; t = 1.729 n = 7, 98% confidence level; t = 3.143 n = 16, 95% confidence level; t = 2.131 n = 23, 99% confidence level; t = 2.819
8.18 Find the ME a. n = 20, 90% confidence level, s = 36 ⎞ = 13.9182 ME = tα 2 s = ME = 1.729 ⎛⎜ 36 ⎟ n 120 ⎠ ⎝ b. n = 7, 98% confidence level, s = 16 ME = tα 2 s = ME = 3.143 ⎛⎜ 16 ⎞⎟ = 19.007 n 7⎠ ⎝ c. n = 16, 99% confidence level, x1 = 15, x2 = 17, x3 = 13, x4 = 11, x5 = 14 x = 14, s = 2.58199 ME = 5.841⎛⎜ 2.58199 ⎞⎟ = 7.5407 4⎠ ⎝
Chapter 8: Estimation: Single Population
8.19 Time spent driving to work for n = 5 people Descriptive Statistics: Driving_Ex8.19 Variable Sum Driving_Ex8.19 192.00
N
N*
Mean
SE Mean
TrMean
StDev
Variance
CoefVar
5
0
38.40
2.66
*
5.94
35.30
15.47
Variable Driving_Ex8.19
Minimum 30.00
Q1 32.50
Median 40.00
Q3 43.50
Maximum 45.00
Range 15.00
IQR 11.00
a. Calculate the standard error s = 5.94138 2.6571 n 5 b. Find the value of t for the 95% confidence interval tv ,α 2 = 2.776 c. Calculate the width for a 95% confidence interval for the population mean ME = tα 2 s = ME = 2.776 ⎛⎜ 5.94138 ⎞⎟ = 14.752 n 5⎠ ⎝ 8.20 Find the LCL and UCL for each of the following: a. alpha = .05, n = 25, sample mean = 560, s = 45 ⎞ = 560 ± 2.064 ⎛ 45 ⎞ = 541.424 to 578.576 x ± tα 2 ⎛⎜ s ⎟ ⎜ ⎟ n⎠ 25 ⎠ ⎝ ⎝ b. alpha/2 = .05, n = 9, sample mean = 160, sample variance = 36 ⎞ = 160 ± 1.860 ⎛ 6 ⎞ = 156.28 to 163.72 x ± tα 2 ⎛⎜ s ⎟ ⎜ ⎟ n⎠ 9⎠ ⎝ ⎝ c. 1 − α = .98, n = 22, sample mean = 58, s = 15 ⎞ = 58 ± 2.518 ⎛15 ⎞ = 49.9474 to 66.0526 x ± tα 2 ⎛⎜ s ⎜ ⎟ ⎟ n 22 ⎝ ⎠ ⎝ ⎠ 8.21 Calculate the margin of error to estimate the population mean for: a. 98% confidence level; n = 64, sample variance = 144 ⎞ = 3.585 ME = tα 2 s = ME = 2.390 ⎛⎜12 ⎟ 64 n ⎝ ⎠ b. 99% confidence level; n = 120, sample standard deviation = 100 ⎞ = 24.2824 ME = tα 2 s = ME = 2.660 ⎛⎜100 ⎟ n 120 ⎠ ⎝ c. 95% confidence level; n = 200, sample standard deviation = 40 ⎞ = 5.65685 ME = tα 2 s = ME = 2.000 ⎛⎜ 40 ⎟ n 200 ⎠ ⎝
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8.22 Calculate the width for each of the following: a. alpha = 0.05, n = 6, s = 40 = 2 ⋅ 2.571⎛⎜ 40 ⎞⎟ = 2(41.98425) = 83.9685 w = 2 ME = 2t α 2 s 6⎠ n ⎝ b. alpha = 0.01, n = 22, sample variance = 400 ⎞ = 2(12.07142) = 24.1428 = 2 ⋅ 2.831⎛⎜ 20 w = 2 ME = 2t α 2 s ⎟ 22 n ⎝ ⎠ c. alpha = 0.10, n = 25, s = 50 ⎞ = 2(17.11) = 34.22 = 2 ⋅ 1.711⎛⎜ 50 w = 2 ME = 2t α 2 s ⎟ 25 ⎠ n ⎝ 8.23 a. 95% confidence interval: Results for: TOC.xls One-Sample T: Leak Rates (cc/sec.) Variable Leak Rates (
N 50
Mean 0.05150
StDev 0.02186
SE Mean 0.00309
95.0% CI ( 0.04529, 0.05771)
b. 98% confidence interval: One-Sample T: Leak Rates (cc/sec.) Variable Leak Rates (
N 50
Mean 0.05150
StDev 0.02186
SE Mean 0.00309
98.0% CI ( 0.04406, 0.05894)
8.24 a. Results for: Sugar.xls Descriptive Statistics: Weights Variable Mean Weights 0.95 Variable Weights
N
Mean
Median
TrMean
StDev
100
520.95
518.75
520.52
9.45
Minimum 504.70
Maximum 544.80
Q1 513.80
Q3 527.28
90% confidence interval:
SE
Results for: Sugar.xls One-Sample T: Weights Variable Weights
N 100
Mean 520.948
StDev 9.451
SE Mean 0.945
90.0% CI ( 519.379, 522.517)
b. narrower since a smaller value of z will be used in generating the 80% confidence interval. 8.25
n = 9, x = 157.82, s = 38.89, t8,.025 = 2.306
margin of error: ± 2.306(38.89/3) = ± 29.8934 8.26
n = 7, x = 74.7143, s = 6.3957, t6,.025 = 2.447
margin of error: ± 2.447(6.3957/ 7 ) = ± 5.9152
Chapter 8: Estimation: Single Population
8.27
a. n = 10, x = 15.90, s = 5.30, t9,.005 = 3.25 15.90 ± 3.25(5.30/ 10 ) = 10.453 up to 21.347 b. narrower since the t-score will be smaller for a 90% confidence interval than for a 99% confidence interval
8.28
n = 25, x = 42, 740, s = 4, 780, t24,.05 = 1.711 42,740 ± 1.711(4780/5) = $41,104.28 up to $44,375.72
8.29
n = 9, x = 16.222, s = 4.790, t8,.10 = 1.86 We must assume a normally distributed population 16.222 ± 1.86(4.790/3) = 13.252 up to 19.192
8.30 Find the standard error of the proportion for pˆ (1 − pˆ ) .3(.7) = .02898 a. n = 250, pˆ = 0.3 = n 250 pˆ (1 − pˆ ) .45(.55) = .03761 = b. n = 175, pˆ = 0.45 n 175 pˆ (1 − pˆ ) .05(.95) = .010897 = c. n = 400, pˆ = 0.05 n 400 8.31 Find the margin of error for a. n = 250, pˆ = 0.3, α = .05 zα 2 b. n = 175, pˆ = 0.45, α = .08 zα 2 c. n = 400, pˆ = 0.05, α = .04 zα 2
.3(.7) pˆ (1 − pˆ ) = .056806 = 1.96 250 n .45(.55) pˆ (1 − pˆ ) = .05810 = 1.75 175 n .05(.95) pˆ (1 − pˆ ) = .02234 = 2.05 400 n
8.32 Find the confidence level for estimating the population proportion for a. 92.5% confidence level; n = 650, pˆ = .10 pˆ (1 − pˆ ) .10(1 − .10) = .10 ± 1.78 = .079055 to .120945 n 650 b. 99% confidence level; n = 140, pˆ = .01 pˆ ± zα 2
pˆ (1 − pˆ ) .01(1 − .01) = .01 ± 2.58 = 0.0 to .031696 n 140 c. alpha = .09; n = 365, pˆ = .50 pˆ ± zα 2
pˆ ± zα 2
pˆ (1 − pˆ ) .5(.5) = .50 ± 1.70 = .4555 to .5445 n 650
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8.33 n = 142, 87 answered GMAT or GRE is ‘very important’. 95% confidence interval for the population proportion: pˆ (1 − pˆ ) .61268(1 − .61268) = .61268 = 1.96 = .53255 to .6928 pˆ = zα 2 n 142 8.34
n = 95,
pˆ = 67 / 95 = .7053, z.005 = 2.58
pˆ (1 − pˆ ) .7053(.2947) = .7053 ± (2.58) = 95 n 99% confidence interval: .5846 up to .8260 pˆ ± zα / 2
8.35 a. unbiased point estimate of proportion: Tally for Discrete Variables: Adequate Variety Adequate Variety 1 2 N=
Count CumCnt 135 135 221 356 356
Percent CumPct 37.92 37.92 62.08 100.00
x 135 = = .3792 n 356 b. 90% confidence interval: n = 356, pˆ = 135 / 356 = .3792, z.05 = 1.645 pˆ =
pˆ (1 − pˆ ) = .3792 ± (1.645) .3792(.6208) / 356 = n .3369 up to .4215 pˆ ± zα / 2
8.36
n = 320,
pˆ = 80 / 320 = .25, z.025 = 1.96
pˆ (1 − pˆ ) = .25 ± (1.96) .25(.75) / 320 = n 95% confidence interval: .2026 up to .2974 pˆ ± zα / 2
8.37
n = 400,
pˆ = 320 / 400 = .80, z.01 = 2.326
pˆ (1 − pˆ ) .80(1 − .80) = .80 − 2.326 = .75348 400 n b. width of a 90% confidence interval .8(1 − .8) pˆ (1 − pˆ ) = 2 ⋅ 1.645 = 2(.0329) = .0658 w = 2 ME = 2 zα 2 400 n
a. LCL = pˆ − zα / 2
Chapter 8: Estimation: Single Population
8.38
width = .545-.445 = .100; ME = 0.05 pˆ = 0.495 .495(.505) pˆ (1 − pˆ ) = = .0355 198 n .05 = zα / 2 (.0355), zα / 2 = 1.41 α = 2[1 − Fz (1.41)] = .0793 100(1-.1586)% = 84.14%
8.39
n = 420,
pˆ = 223 / 420 = .5310, z.025 = 1.96
pˆ (1 − pˆ ) = .5310 ± (1.96) .5310(.4690) / 420 = n 95% confidence interval: .4833 up to .5787 The margin of error is .0477 pˆ ± zα / 2
8.40
n = 246,
pˆ = 40 / 246 = .1626, z.01 = 2.326
pˆ (1 − pˆ ) = .1626 ± (2.326) .1626(.8374) / 246 = n 98% confidence interval: .1079 up to .2173 pˆ ± zα / 2
8.41
a. n = 246,
pˆ = 232 / 246 = .9431, z.01 = 2.326
pˆ (1 − pˆ ) = .9431 ± (2.326) .9431(.0569) / 246 = n 98% confidence interval: .9087 up to .9775 b. n = 246, pˆ = 10 / 246 = .0407, z.01 = 2.326 pˆ ± zα / 2
pˆ (1 − pˆ ) = .0407 ± (2.326) .0407(.9593) / 246 = n 98% confidence interval: .0114 up to .0699 pˆ ± zα / 2
8.42 a. n = 10, x = 257, s = 37.2, t9,.05 = 1.833 ⎞ = 257 ± 1.833(37.2/ 10 ) = 235.4318 up to 278.5628 x ± tα 2 ⎛⎜ s ⎟ n⎠ ⎝ assume that the population is normally distributed b. 95% and 98% confidence intervals: ⎞ = 257 ± 2.262(37.2/ 10 ) = 230.39 up to 283.61 [95%]: x ± tα 2 ⎛⎜ s ⎟ n⎠ ⎝ ⎞ = 257 ± 2.821(37.2/ 10 ) = 223.815 up to [98%]: x ± tα 2 ⎛⎜ s ⎟ n⎠ ⎝ 290.185
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8.43
n = 16, x = 150, s = 12, t15,.025 = 2.131 ⎞ = 150 ± 2.131(12/4) = 143.607 up to 156.393 x ± tα 2 ⎛⎜ s ⎟ n⎠ ⎝ It is recommended that he stock 157 gallons.
8.44
n = 50, x = 30, s = 4.2, z.05 = 1.645 = 30 ± 1.645(4.2/ 50 ) = 29.0229 up to 30.9771
8.45
Results from Minitab:
Descriptive Statistics: Passengers8_45 Variable Mean Passenge 3.46 Variable Passenge
N
Mean
Median
TrMean
StDev
50
136.22
141.00
136.75
24.44
Minimum 86.00
Maximum 180.00
Q1 118.50
Q3 152.00
One-Sample T: Passengers8_45
SE
Variable Passengers8_
N 50
Mean 136.22
StDev 24.44
SE Mean 3.46
(
95.0% CI 129.27, 143.17)
95% confidence interval: 129.27 up to 143.17 8.46
a. Use a 5% risk. Incorrect labels = 8/48 = 0.1667 0.1667(0.8333) 0.1667 ± 1.96 = 0.1667 ± 0.1054 = 0.0613 up to 0.2721 48 b. For a 90% confidence interval, 0.1667(0.8333) 0.1667 ± 1.645 = 0.1667 ± 0.0885 = 0.0782 up to 48 0.2552
8.47
a. The minimum variance unbiased point estimator of the population ∑ X i = 27 = 3.375. The unbiased point mean is the sample mean: X = 8 n 2 xi − nx 2 94.62 − 8(3.375) 2 ∑ 2 estimate of the variance: s = = = .4993 n −1 7 x 3 b. pˆ = = = .375 n 8
8.48 3.69 – 3.59 = 0.10 = zα / 2 (1.045 / 457), zα / 2 = 2.05 α = 2[1 − Fz (2.05)] = .0404 100(1-.0404)% = 95.96%
Chapter 8: Estimation: Single Population
8.49
n = 174, x = 6.06, s = 1.43
6.16 – 6.06 = .1 = zα / 2 (1.43 / 174), zα / 2 = .922 α = 2[1 − Fz (.92)] = .3576 100(1-.3576)% = 64.24% 8.50 n = 33 accounting students who recorded study time a. An unbiased, consistent, and efficient estimator of the population mean is the sample mean x = 8.545 b. Find the sampling error for a 95% confidence interval; Using degrees of freedom = 30, ⎞ = 1.3568 ME = 2.042 ⎛⎜ 3.817 ⎟ 33 ⎝ ⎠ 8.51 n = 25 patient records – the average length of stay is 6 days with a standard deviation of 1.8 days a. find the reliability factor for a 95% interval estimate tα / 2 = 2.064 b. Find the LCL for a 99% confidence interval estimate of the population mean ⎞ . The LCL = 5.257 = 6 − 2.064 ⎛⎜1.8 ME = tα 2 s ⎟ 25 n ⎝ ⎠ 8.52 n = 250, x = 100 a. Find the standard error to estimate population proportion of first timers pˆ (1 − pˆ ) .4(1 − .4) = .03098 = n 250 b. Find the sampling error. Since no confidence level is specified, we find the sampling error (Margin of Error) for a 95% confidence interval. ME = 1.96 (0.03098) = 0.0607 c. For a 92% confidence interval, ME = 1.75 (0.03098) = 0.05422 0.40 ± .05422 giving 0.3457 up to 0.4542 8.53
a. 90% confidence interval reliability factor = tα / 2 = 1.729 b. Find the LCL for a 99% confidence interval LCL = 60.75 – 2.861 21.83159 = 46.78 or approximately 47 20 passengers.
8.54 a. Find a 95% confidence interval estimate for the population proportion of students who would like supplements in their smoothies.
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Tally for Discrete Variables: Supplements, Health Consciousness Supplements No 0 Yes 1 N=
n = 113,
Count 42 71 113
Percent 37.17 62.83
Health Consciousness Very 1 Moderately 2 Slight 3 Not Very 4 N=
Count 29 55 20 9 113
Percent 25.66 48.67 17.70 7.96
pˆ = 71/113 = .62832, z.05 = 1.96
pˆ (1 − pˆ ) .62832(1 − .62832) = .62832 ± 1.96 n 113 0.0891 = 0.5392 up to 0.71742. pˆ ± zα / 2
= 0.62832 ±
b. pˆ = 29 / 113 = 0.2566 For 98% confidence level, pˆ ± 2.33
(0.2566)(1 − 0.2566) = 0.2566 ± 0.09573 113
or 0.1609 up to
0.3523 c. pˆ = 77 / 113 = 0.6814 0.6814 ± 1.645
(0.6814)(1 − 0.6814) = 0.6814 ± 0.0721 or 0.6093 up to 113
0.7535 8.55
n = 100 students at a small university. a. Estimate the population grade point average with 95% confidence level One-Sample T: GPA Variable GPA
N 100
Mean 3.12800
StDev 0.36184
SE Mean 0.03618
95% CI (3.05620, 3.19980)
b. Estimate the population proportion of students who were very dissatisfied (code 1) or moderately dissatisfied (code 2) with parking. Use a 90% confidence level. Tally for Discrete Variables: Parking Parking 1 2 3 4 5 N=
Count 19 26 18 18 19 100
n = 100, .45 ± 1.645
Percent 19.00 26.00 18.00 18.00 19.00
pˆ = 45 /100 = .45, z.05 = 1.645 , pˆ ± zα / 2
pˆ (1 − pˆ ) n
.45(1 − .45) = .45 ± .08184 = .368162 up to .53184 100
c. Estimate the proportion of students who were at least moderately satisfied (codes 4 and 5) with on-campus food service
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Tally for Discrete Variables: Dining Dining 1 2 3 4 5 N=
Count 14 26 21 20 19 100
Percent 14.00 26.00 21.00 20.00 19.00
n = 100,
pˆ = 39 /100 = .39, z.05 = 1.645
pˆ ± zα / 2
pˆ (1 − pˆ ) .39(1 − .39) = .39 ± 1.645 = .39 ± .08023 = .30977 up 100 n
to .47023. 8.56
a. Estimate the average age of the store’s customers by the sample mean ∑ xi = 6310 = 50.48 x= n 125 To find a confidence interval estimate we will assume a 95% confidence level: 13.06 50.48 ± 1.96 = 50.48 ± 2.29 ; 48.19 up to 52.77 years 125 b. Estimate the population proportion of customers dissatisfied with the delivery system Tally for Discrete Variables: Dissatisfied with Delivery Dissatisfied with Delivery 1 2 N=
Count 9 116 125
Percent 7.20 92.80
pˆ = 9 /125 = .072 ; Assuming a 95% confidence level, we find: 0.072 ± 1.96
(0.072)(1 − 0.072) = 0.072 ± 0.0453 or 0.0267 up to 0.1173 125
c. Estimate the population mean amount charged to a Visa credit card Descriptive Statistics: Cost of Flowers Variable Cost of Flowers
Method of Payment American Express Cash Master Card Other Visa
Mean 52.99 51.34 54.58 53.42 52.65
SE Mean 2.23 4.05 3.11 2.99 2.04
TrMean 52.83 51.46 54.43 53.72 52.58
StDev 10.68 16.19 15.25 14.33 12.71
The population mean can be estimated by the sample mean amount charged to a Visa credit card = $52.65. 8.57 n = 500 motor vehicle registrations, 200 were mailed, 160 paid in person, remainder paid on-line. a. Estimate the population proportion to pay for vehicle registration renewals in person, use a 90% confidence level.
Median 50.55 50.55 55.49 54.85 50.65
180
th
Statistics for Business & Economics, 6 edition
Test and CI for One Proportion Test of p = 0.5 vs p not = 0.5 Sample X N Sample p 90% CI 1 160 500 0.320000 (0.285686, 0.354314)
The 90% confidence interval is from 28.56856% up to 35.4314% b. Estimate the population proportion of on-line renewals, use 95% confidence. Test and CI for One Proportion Test of p = 0.5 vs p not = 0.5 Sample X N Sample p 95% CI 1 140 500 0.280000 (0.240644, 0.319356)
The 95% confidence interval is from 24.0644% up to 31.9356% 8.58
From the data in 8.57, find the confidence level if the interval extends from 0.34 up to 0.46. ME = ½ the width of the confidence interval. 0.46 – 0.34 = 0.12 / 2 = 0.06 pˆ (1 − pˆ ) or ME = z α 2 n 0.06 = z α 2
(0.4)(0.6) 500 Solving for z:
z α 2 = 2.74
Area from the z-table = .4969 x 2 = .9938. The confidence level is 99.38% 8.59 From the data in 8.57, find the confidence level if the interval extends from 23.7% up to 32.3%. ME = ½ the width of the confidence interval. .323 – .237 = .086 / 2 = .043 and pˆ = .28 pˆ (1 − pˆ ) .28(1 − .28) .043 = zα 2 solving for z: zα 2 = 2.14 n 500 Area from the z-table = .4838 x 2 = .9676. The confidence level is 96.76%
ME = zα 2
8.60
a. What is the margin of error for a 99% confidence interval pˆ (1 − pˆ ) ME = zα 2 pˆ = x = 250 = .7143 , n 350 n .7143(1 − .7143) 350 ME = .0623
ME = 2.58
=
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181
b. Is the margin of error for a 95% confidence larger, smaller or the same as the 99% confidence level? The margin of error will be smaller (more precise) for a lower confidence level. The difference in the equation is the value for z which would drop from 2.58 down to 1.96. 8.61 Compute the 98% confidence interval of the mean age of on-line renewal users. n= 460, sample mean = 42.6, s = 5.4. ⎞ = 42.6 ± .58664 = 42.0134 up to = 42.6 ± 2.33 ⎛⎜ 5.4 x ± tα 2 s ⎟ 460 ⎠ n ⎝ 43.18664
