Chapter 8: Estimation: Single Population 1
Chapter 8:
Estimation: Single Population
8.1 a. Check for nonnormality
The distribution shows no significant evidence of nonnormality.
b. Point estimate of the population mean that is unbiased, efficient and consistent.
Unbiased point estimator is the sample mean: = 7.0
- The unbiased point estimate of the variance of the sample mean:
8.2a. Evidence of non-normality?
No evidence of non-normality.
b. The minimum variance unbiased point estimator of the population mean is the sample mean: = 101.375
- The unbiased point estimate of the variance of the sample mean:
d.
8.3 n = 10 economists forecast for percentage growth in real GDP in the next year
Descriptive Statistics: RGDP_Ex8.3
Variable N N* Mean SE Mean TrMean StDev Variance CoefVar Sum
RGDP_Ex8.3 10 0 2.5700 0.0716 2.5625 0.2263 0.0512 8.81 25.7000
Variable Minimum Q1 Median Q3 Maximum Range IQR
RGDP_Ex8.3 2.2000 2.4000 2.5500 2.7250 3.0000 0.8000 0.3250
a. Unbiased point estimator of the population mean is the sample mean: 2.57
b. The unbiased point estimate of the population variance:
c. Unbiased point estimate of the variance of the sample mean
d. Unbiased estimate of the population proportion:
e. Unbiased estimate of the variance of the sample proportion:
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: 24.42
b. The unbiased point estimate of the population variance:
c. Unbiased point estimate of the variance of the sample mean
d. Unbiased estimate of the population proportion:
e. Unbiased estimate of the variance of the sample proportion:
8.5a. Check each variable for normal distribution:
No evidence of non-normality in Meals, Attendance or Ages
- Unbiased estimates of population mean and variance:
Descriptive Statistics: Meals, Attendance, Ages
Variable N Mean Median TrMean StDev SE Mean
Meals 30 50.100 50.000 50.192 2.468 0.451
Attendan 25 21.240 21.000 21.348 2.505 0.501
Ages 25 12.240 12.000 12.217 1.964 0.393
Variable Minimum Maximum Q1 Q3
Meals 45.000 55.000 48.000 52.000
Attendan 15.000 25.000 19.500 23.500
Ages 9.000 16.000 10.000 14.000
Variable Unbiased estimate of mean Unbiased estimate of variance (s2)
Meals50.100(2.468)2 = 6.0910
Attendance21.240(2.505)2 = 6.2750
Ages12.240(1.964)2 = 3.8573
8.6a.
b.
is most efficient since
- Relative efficiency between Y and :
Relative efficiency between Z and :
8.7a. Evidence of non-normality?
No evidence of nonnormality exists.
b. The minimum variance unbiased point estimator of the population mean is the sample mean: .0515
- The unbiased point estimate of the variance of the sample mean:
8.8a. Evidence of non-normality?
No evidence of the data distribution coming from a non-normal population
- The minimum variance unbiased point estimator of the population mean is the sample mean: 3.8079
Descriptive Statistics: Volumes
Variable N Mean Median TrMean StDev SE Mean
Volumes 75 3.8079 3.7900 3.8054 0.1024 0.0118
Variable Minimum Maximum Q1 Q3
Volumes 3.5700 4.1100 3.7400 3.8700
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: = +/- 2.05
b. 88% confidence level: = +/- 1.56
c. 85% confidence level: = +/- 1.44
d. = +/- 1.81
e. = .07 = +/- 1.48
8.10Calculate the margin of error to estimate the population mean
- 98% confidence level; n = 64, variance = 144
= = 3.495
b. 99% confidence interval, n=120; standard deviation = 100
= = 23.552
8.11Calculate the width to estimate the population mean, for
- 90% confidence level, n = 100, variance = 169
width = 2ME = = = 4.277
- 95% confidence interval, n = 120, standard deviation = 25
width = 2ME = = = 8.9461
8.12Calculate the LCL and UCL
a. = = 40.2 to 59.8
b. = = 81.56 to 88.44
c. = = 506.2652 to 513.73478
8.13 a.
187.9 1.28(32.4/3) = 174.076 up to 201.724
b. 210.0 – 187.9 = 22.1 =
100(1-.0404)% = 95.96%
8.14 a. Find the reliability factor for 92% confidence level: = +/- 1.75
b. Calculate the standard error of the mean
= .63246
c. Calculate the width of the 92% confidence interval for the population mean
width = 2ME = = = 2.2136
8.15 a.
= 2.90 1.96(.45/5) = 2.7236 up to 3.0764
b. 2.99 – 2.90 = .09 =
100(1-.3174)% = 68.26%
8.16 a.
4.07 2.58(.12/4) = 3.9926 up to 4.1474
- narrower since the z score for a 95% confidence interval is smaller than the z score for the 99% confidence interval
- narrower due to the smaller standard error
- wider due to the larger standard error
8.17 Find the reliability factor
- 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
- n = 20, 90% confidence level, s = 36
= = 13.9182
- n = 7, 98% confidence level, s = 16
= = 19.007
c. n = 16, 99% confidence level, x1 = 15, x2 = 17, x3 = 13, x4 = 11, x5 = 14
= 7.5407
8.19 Time spent driving to work for n = 5 people
Descriptive Statistics: Driving_Ex8.19
Variable N N* Mean SE Mean TrMean StDev Variance CoefVar Sum
Driving_Ex8.19 5 0 38.40 2.66 * 5.94 35.30 15.47 192.00
Variable Minimum Q1 Median Q3 Maximum Range IQR
Driving_Ex8.19 30.00 32.50 40.00 43.50 45.00 15.00 11.00
- Calculate the standard error
2.6571
b. Find the value of t for the 95% confidence interval
= 2.776
- Calculate the width for a 95% confidence interval for the population mean
= = 14.752
8.20 Find the LCL and UCL for each of the following:
a. alpha = .05, n = 25, sample mean = 560, s = 45
= = 541.424 to 578.576
b. alpha/2 = .05, n = 9, sample mean = 160, sample variance = 36
= = 156.28 to 163.72
c. = .98, n = 22, sample mean = 58, s = 15
= = 49.9474 to 66.0526
8.21 Calculate the margin of error to estimate the population mean for:
- 98% confidence level; n = 64, sample variance = 144
= = 3.585
- 99% confidence level; n = 120, sample standard deviation = 100
= = 24.2824
- 95% confidence level; n = 200, sample standard deviation = 40
= = 5.65685
8.22 Calculate the width for each of the following:
a. alpha = 0.05, n = 6, s = 40
b. alpha = 0.01, n = 22, sample variance = 400
c. alpha = 0.10, n = 25, s = 50
8.23 a. 95% confidence interval:
Results for: TOC.xls
One-Sample T: Leak Rates (cc/sec.)
Variable N Mean StDev SE Mean 95.0% CI
Leak Rates ( 50 0.05150 0.02186 0.00309 ( 0.04529, 0.05771)
b. 98% confidence interval:
One-Sample T: Leak Rates (cc/sec.)
Variable N Mean StDev SE Mean 98.0% CI
Leak Rates ( 50 0.05150 0.02186 0.00309 ( 0.04406, 0.05894)
8.24 a.
Results for: Sugar.xls
Descriptive Statistics: Weights
Variable N Mean Median TrMean StDev SE Mean
Weights 100 520.95 518.75 520.52 9.45 0.95
Variable Minimum Maximum Q1 Q3
Weights 504.70 544.80 513.80 527.28
90% confidence interval:
Results for: Sugar.xls
One-Sample T: Weights
Variable N Mean StDev SE Mean 90.0% CI
Weights 100 520.948 9.451 0.945 ( 519.379, 522.517)
b. narrower since a smaller value of z will be used in generating the 80%
confidence interval.
8.25
margin of error: 2.306(38.89/3) = 29.8934
8.26
margin of error: 2.447(6.3957/) = 5.9152
8.27 a.
15.90 3.25(5.30/) = 10.453 up to 21.347
- narrower since the t-score will be smaller for a 90% confidence interval than for a 99% confidence interval
8.28
42,740 1.711(4780/5) = $41,104.28 up to $44,375.72
8.29
We must assume a normally distributed population
16.222 1.86(4.790/3) = 13.252 up to 19.192
8.30Find the standard error of the proportion for
- n = 250, = 0.3 = .02898
- n = 175, = 0.45 = .03761
- n = 400, = 0.05 = .010897
8.31Find the margin of error for
- n = 250, = 0.3, α = .05 = .056806
- n = 175, = 0.45, α = .08 = .05810
- n = 400, = 0.05, α = .04 = .02234
8.32 Find the confidence level for estimating the population proportion for
a. 92.5% confidence level; n = 650, = .10
= = .079055 to .120945
b. 99% confidence level; n = 140, = .01
= = 0.0 to .031696
c. alpha = .09; n = 365, = .50
= = .4555 to .5445
8.33n = 142, 87 answered GMAT or GRE is ‘very important’. 95% confidence interval for the population proportion:
= = .53255 to .6928
8.34
= .7053 (2.58) =
99% confidence interval: .5846 up to .8260
8.35 a. unbiased point estimate of proportion:
Tally for Discrete Variables: Adequate Variety
Adequate Variety Count CumCnt Percent CumPct
1 135 135 37.92 37.92
2 221 356 62.08 100.00
N= 356
b. 90% confidence interval:
= .3792 (1.645) =
.3369 up to .4215
8.36
= .25 (1.96) =
95% confidence interval: .2026 up to .2974
8.37
a. LCL = = = .75348
b. width of a 90% confidence interval
8.38 width = .545-.445 = .100; ME = 0.05
.05 = (.0355),
100(1-.1586)% = 84.14%
8.39
= .5310 (1.96) =
95% confidence interval: .4833 up to .5787
The margin of error is .0477
8.40
= .1626 (2.326) =
98% confidence interval: .1079 up to .2173
8.41 a.
= .9431 (2.326) =
98% confidence interval: .9087 up to .9775
b.
= .0407 (2.326) =
98% confidence interval: .0114 up to .0699
8.42 a.
= 257 1.833(37.2/) = 235.4318 up to 278.5628
assume that the population is normally distributed
- 95% and 98% confidence intervals:
[95%]: = 257 2.262(37.2/) = 230.39 up to 283.61
[98%]: = 257 2.821(37.2/) = 223.815 up to 290.185
8.43
= 150 2.131(12/4) = 143.607 up to 156.393
It is recommended that he stock 157 gallons.
8.44
= 30 1.645(4.2/) = 29.0229 up to 30.9771
8.45 Results from Minitab:
Descriptive Statistics: Passengers8_45
Variable N Mean Median TrMean StDev SE Mean
Passenge 50 136.22 141.00 136.75 24.44 3.46
Variable Minimum Maximum Q1 Q3
Passenge 86.00 180.00 118.50 152.00
One-Sample T: Passengers8_45
Variable N Mean StDev SE Mean 95.0% CI
Passengers8_ 50 136.22 24.44 3.46 ( 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
= 0.0613 up to 0.2721
b. For a 90% confidence interval,
= 0.0782 up to 0.2552
8.47a. The minimum variance unbiased point estimator of the population mean is the sample mean: 3.375. The unbiased point estimate of the variance:
b.
8.48 3.69 – 3.59 = 0.10 =
100(1-.0404)% = 95.96%
8.49
6.16 – 6.06 = .1 =
100(1-.3576)% = 64.24%
8.50n = 33 accounting students who recorded study time
- An unbiased, consistent, and efficient estimator of the population mean is the sample mean = 8.545
- Find the sampling error for a 95% confidence interval; Using degrees of freedom = 30,
= 1.3568
8.51n = 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
b. Find the LCL for a 99% confidence interval estimate of the population mean
= . The LCL = 5.257
8.52 n = 250, x = 100
- Find the standard error to estimate population proportion of first timers
= .03098
- 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 =
b. Find the LCL for a 99% confidence interval
LCL = 60.75 – 2.861 = 46.78 or approximately 47 passengers.
8.54 a. Find a 95% confidence interval estimate for the population proportion of students who would like supplements in their smoothies.
Tally for Discrete Variables: Supplements, Health Consciousness
Health
Supplements Count Percent Consciousness Count Percent
No 0 42 37.17 Very 1 29 25.66
Yes 1 71 62.83 Moderately 2 55 48.67
N= 113 Slight 3 20 17.70
Not Very 4 9 7.96
N= 113
= = 0.62832 ± 0.0891 = 0.5392 up to 0.71742.
b. For 98% confidence level,
or 0.1609 up to 0.3523
c.
or 0.6093 up to 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 N Mean StDev SE Mean 95% CI
GPA 100 3.12800 0.36184 0.03618 (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 Count Percent
1 19 19.00
2 26 26.00
3 18 18.00
4 18 18.00
5 19 19.00
N= 100
, = .45 ± .08184 = .368162 up to .53184
- Estimate the proportion of students who were at least moderately satisfied
(codes 4 and 5) with on-campus food service
Tally for Discrete Variables: Dining
Dining Count Percent
1 14 14.00
2 26 26.00
3 21 21.00
4 20 20.00
5 19 19.00
N= 100
= = .39 ± .08023 = .30977 up to .47023.
8.56 a. Estimate the average age of the store’s customers by the sample mean
To find a confidence interval estimate we will assume a 95% confidence level:
; 48.19 up to 52.77 years
b. Estimate the population proportion of customers dissatisfied with the delivery system
Tally for Discrete Variables: Dissatisfied with Delivery
Dissatisfied
with
Delivery Count Percent
1 9 7.20
2 116 92.80
N= 125
; Assuming a 95% confidence level, we find:
or 0.0267 up to 0.1173
- Estimate the population mean amount charged to a Visa credit card
Descriptive Statistics: Cost of Flowers
Method of
Variable Payment Mean SE Mean TrMean StDev Median
Cost of Flowers American Express 52.99 2.23 52.83 10.68 50.55
Cash 51.34 4.05 51.46 16.19 50.55
Master Card 54.58 3.11 54.43 15.25 55.49
Other 53.42 2.99 53.72 14.33 54.85
Visa 52.65 2.04 52.58 12.71 50.65
The population mean can be estimated by the sample mean amount charged to a Visa credit card = $52.65.
8.57n = 500 motor vehicle registrations, 200 were mailed, 160 paid in person, remainder paid on-line.
- Estimate the population proportion to pay for vehicle registration renewals in person, use a 90% confidence level.
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%
8.58From 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
or
Solving for z:
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
solving for z:
Area from the z-table = .4838 x 2 = .9676. The confidence level is 96.76%
8.60a. What is the margin of error for a 99% confidence interval
, =
ME = .0623
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 43.18664