CHAPTER 5
SAMPLING AND SAMPLING DISTRIBUTIONS
5.1. Parameters are numerical measures of populations. Sample statistics are numerical measures ofsamples. An estimator is a sample statistic used for estimating a population parameter. 5-2. x = 97.9225 (estimate of )
s = 51.8303 (estimate of )
s2 = 2,686.38 (estimate of 2—the population variance) 5-3. pˆ = x/n = 5/12 = 0.41667
(5 out of 12 accounts are over $100.) 5-4. x = 2121.667 s = 1737.714
Basic Statistics from Raw Data
Measures of Central tendency
Mean 2121.6667 Median Measures of Dispersion If the data is of a Sample Population Variance 3019651.52 St. Dev. 1737.71445
5.5. average price = 4.367 standard deviation = 0.3486 Basic Statistics from Raw Data
Measures of Central tendency
Mean 4.3676471 Median
5.6. pˆ = x/n = 11/18 = 0.6111, where x = the number of users of the product.
5.7. We need 25 elements from a population of 950 elements. Use the rows of Table 5-1, the rightmost 3 digits of each group starting in row 1 (left to right). So we skip any such 3-digit number that is either > 950 or that has been generated earlier in this list, giving us a list of 25 different numbers in the desired range. The chosen numbers are:
480, 11, 536, 647, 646, 179, 194, 368, 573, 595, 393, 198, 402, 130, 360, 527, 265, 809, 830, 167, 93, 243, 680, 856, 376.
5.8. We will use again Table 5-1, using columns this time. We will use right-hand columns, first 4 digits from the right (going down the column):
4,194 3,402 4,830 3,537 1,305.
5.9 We will use Table 5-1, sets of 2 columns using all 5 digits from column 1 and the first 3 digits from column 2, continuing by reading down in these columns. Then we will continue to the set: column 3 and first 3 digits column 4. We skip any numbers that are > 40,000,000. The resulting voter numbers are:
10,480,150 22,368,465 24,130,483 37,570,399 1,536,020.
5.10.There are 7 x 24 x 60 minutes in one week: (7)(24)(60) = 10,080 minutes. We will use Table 5-1 Start in the first row and go across the row, then to the next row (left to right using all 5 digits in each set), discarding any of the resulting 5-digit numbers that are > 10,080. The resulting minute numbers are:
1,536 2,011 6,243 7,856 6,121 6,907
5-11. A sampling distribution is the probability distribution of a sample statistic. The sampling distribution is useful in determining the accuracy of estimation results.
5.12.Only if the population is itself normal.
5-13. E
X = 125 SE
X / n 20/ 5 = 8.9445.14.The fact that, in the limit, the population distribution does not matter. Thus the theorem is very general.
5.17. Pˆ is binomial. Since np = 1.2, the Central Limit Theorem does not apply and we cannot use the normal distribution. 5.18. = 1,247 2 = 10,000 n = 100 P( X < 1,230) = P
10 / 100 247 , 1 230 , 1 Z = P(Z < –1.7) = .5 – .4554 = 0.0446 Sampling Distribution of Sample MeanPopulation Distribution
Mean Stdev
1247 100
Sample Size Sampling Distribution of X-bar
n 100 Mean Stdev 1247 10 P(X<x) x 0.0446 1230 5.19.P
X 8
= 1 – P
X 8
= 1 – P(–8 < X < 8) = 1 – P
150 / 55 8 150 / 55 8 Z = 1 – P(–1.78 < Z < 1.78) = 1 – 2(.4625) = 0.075 5.20.P(X > 3.6) = P
100 / 5 . 1 4 . 3 6 . 3 Z = P(Z > 1.333) = 0.0912 Sampling Distribution of Sample MeanPopulation Distribution
Mean Stdev
3.4 1.5
Sample Size Sampling Distribution of X-bar
n 100 Mean Stdev
3.4 0.15
x P(X>x)
5.21.P(12 < X < 15) = P 36 / 2 . 1 1 . 13 15 36 / 2 . 1 1 . 13 12 Z = P(–5.5 < Z < 9.5) = 2 (.5) = 1.000 (approximately) (Use template: Sampling Distribution.xls, sheet: x-bar) Sampling Distribution of Sample Mean
Population Distribution
Mean Stdev Is the populationnormal?
13.1 1.2
Sample Size Sampling Distribution of X-bar
n 36 Mean Stdev 13.1 0.2 x1 P(x1<X<x2) x2 12 1.0000 15 5.22.s = 4,500 n = 225 P
X 800
= P
225 / 500 , 4 800 Z = P
15 / 500 , 4 800 15 / 500 , 4 800 Z = P(–2.667 < Z < 2.667) = 2(.4961) = 0.9923 5-23. p = 0.18 n = 200 P(Pˆ.20) = P 200 / ) 82 )(. 18 (. 18 . 20 . Z = P
02717 . 02 . Z = P(Z .736) = .5 – .2692 = 0.23085.24.The claim is that p = 0.58. We have n = 250 and x / n = 123/250 = 0.492. P(Pˆ .492) = P
250 / ) 42 )(. 58 (. 58 . 492 . Z = P(Z < -2.819) = 0.00245-25. P(X > $3M) = 0.00
Sampling Distribution of Sample Mean
known
Population Distribution
Mean Stdev Is the population normal?
2.6 0.4
Sample Size Sampling Distribution of X-bar
n 75 Mean Stdev 2.6 0.04619 x P(>x) 3 0.0000 5-26. n = 16
= 1.5
= 2 P(X > 0) = P 16 / 2 5 . 1 0 Z = P(Z > -3) = .5 + .4987 = 0.9987 Sampling Distribution of Sample MeanPopulation Distribution
Mean Stdev
1.5 2
Sample Size Sampling Distribution of X-bar
n 16 Mean Stdev
1.5 0.5
x P(X>x)
5.27. p = 1/7 P(Pˆ < .10) = P 180 / ) 7 / 6 )( 7 / 1 ( 143 . 10 . Z = P(Z < 1.648) = 0.5 0.4503 =
0.0497, a low probability. The sample size, along with np and n(1 – p), are large enough here that the sample distribution (over all the different samples of 180 people in the population) of the proportion of people who get hospitalized during the year is going to be pretty close to normal. Therefore, any one such sample proportion will be close to the predicted mean 1/7 with reasonable probability, and 1/10 is far enough away from that mean given our estimated sample standard deviation that the probability of falling even farther away than that from the mean is small. 5-28.
= 700
= 100 n = 60 P(680
X
720) = P 60 / 100 700 720 60 / 100 700 680 Z = 2TA(1.549) = 0.8786 5-29. p =
= 0.35
= (0.35)(0.65)/500 = 0.0213 P
Pˆ p 0.05
= P(Pˆ < 0.30) + P(Pˆ > 0.40)= P 0213 . 0 35 . 0 30 . 0 Z + P 0213 . 0 35 . 0 40 . 0 Z = 1 – 2TA(2.344) = 0.0190
5.30. Estimator B is better. It has a small bias, but its variance is small. This estimator is more likely to produce an estimate that is close to the parameter of interest.
5.31. I would use this estimator because consistency means as n
the probability of getting close to the parameter increases. With a generous budget I can get a large sample size, which will make this probability high.5.32. ˆs2 = 1,287 s 2 = 1 n n 2 ˆs = 99 100 1,287 = 1,300
5.33. Advantage: uses all information in the data.
Disadvantage: may be too sensitive to the influence of outliers. 5.34. Depends also on efficiency and other factors. With respect to the bias:
A has bias = 1/n B has bias = 0.01
A is better than B when 1/n < 0.01, that is, when n > 1/0.01 = 100
5.35. Consistency is important because it means that as you get more data, your probability of getting closer to your “target” increases.
5.36. n = 30, 1 n = 48, 2 n = 32. The three sample means are known. The df for deviations from the 3
three sample means are:
5.37. a) the mean is the best number to use.
mean = 43.667
Deviation Deviation Sample from meansquared
34 -9.667 93.45089 51 7.333 53.77289 40 -3.667 13.44689 38 -5.667 32.11489 47 3.333 11.10889 50 6.333 40.10689 52 8.333 69.43889 44 0.333 0.110889 37 -6.667 44.44889 SSD = 358 degrees of freedom = 8 MSD = SSD / df = 358 / 8 = 44.75
b) choose the means of the respective block of numbers: 40.75, 49.667, 40.5 minimized SSD = 195.917, df = 6, MSD = 32.65283
mean = 40.75 49.667 40.5
Deviation Deviation Sample from meansquared
34 -6.75 45.5625 51 10.25 105.0625 40 -0.75 0.5625 38 -2.75 7.5625 47 -2.667 7.112889 50 0.333 0.110889 52 2.333 5.442889 44 3.5 12.25 37 -3.5 12.25 SSD = 195.9167
c) Each of the numbers themselves. SSD = 0. MSD indicates that the variance is zero, which is true since we are using each of the individual numbers to reduce SSD to zero.
d) SSD = 719, df = 9, MSD = 79.889
mean = 50
Deviation Deviation Sample from meansquared
34 -16 256 51 1 1 40 -10 100 38 -12 144 47 -3 9 50 0 0 52 2 4 44 -6 36 37 -13 169 SSD = 719
5.38. No, because there are n – 1 = 19 – 1 = 18 degrees of freedom for these checks once you know their mean. Since 17 is on less, there is a remaining degree of freedom and you cannot solve for the missing checks.
5.39. Yes. (x + 1 + x + 18 x )/19 = x . Since 18 of the 19 x are known and so is x , we can solve i
the equation for the unknown x .19
5.40. df = n-k
as k increases, df decreases, SSD decreases, MSD decreases 5.41. E(X ) =
= 1,065 V(X ) = 2/n = 5002/100 = 2,500 5.42. 2 = 1,000,000 Want SD(X )
25 SD( X ) = / n = 1,000 / n 1,000 / n
25 n
1,000/25 = 40n
1,600. The sample size must be at least 1,600. 5.43.
= 53
= 10 n = 400Sampling Distribution of Sample Mean
Population Distribution
Mean Stdev
53 10
Sample Size
Sampling Distribution of X-bar
n 400 Mean Stdev 53 0.5 5-44. p = 0.5 n = 120 SE(Pˆ ) = n p p(1 ) = 120 ) 5 )(. 5 (. = 0.0456 5.45.E(Pˆ) = p = 0.2 SE(Pˆ ) = n p p(1 ) = 90 ) 8 )(. 2 (. = 0.04216
5.46.P = 0.5 maximizes the variance of Pˆ. Proof: V(Pˆ ) = n p p(1 ) dp P dV( ˆ) = n 1 dp d (pp 2) = n 1 (1 – 2p) Set the derivative to zero:
n 1
(1 – 2p) = 0 1 = 2p p = 1/2
The assertion may also be demonstrated by trying different values of p. 5.47.P(0.72 < X < 0.82) = P(–10.95 < Z <16 .432) = 2(.5) = 1.00
Sampling Distribution of Sample Mean
known
x1 P(x1<X<x2) x2 0.72 1.0000 0.82 5.48.P( X
1.000) = P 10 / 500 065 , 1 000 , 1 Z = P 500 650 Z = P(Z
1.3) = .5 + .4032 = 0.9032We need to use the Central Limit Theorem for a normal distribution. 5-49.
= 53
= 10 n = 400 P(52 < X < 54) = P 20 / 10 53 54 20 / 10 53 52 Z = P(2 < Z < 2) = 0.9544 5-50. p = 0.5 n = 120 P(Pˆ .45) = P 120 / ) 5 )(. 5 (. 5 . 45 . Z = P(Z
1.095) = 0.8632 5-51. a. $8,128.08 found by $3.3M/406 = 8,128.08 b. P(X < 7000) = P 16 / 2000 08 . 8128 7000 Z = P(Z < 2.256) = .5000 .4880 = 0.012 5.52. 0.06
p
0.10 SE(Pˆ ) = p(1p)/n
0.03 Assume p = 0.06: SE(Pˆ ) = (.06)(.94)/n
.03 (.06)(.94)/n
.032 62.66
n P(X<x) x 0.0120 7000Now, we also know that the function SE(Pˆ ) does not have a maximum point between p = 0.06 and p = 0.10 because the only maximum point of the function occurs at p = 0.5 (as we know from Problem 5-46). Hence SE(Pˆ ) is monotonic between p = 0.06 and 0.10, and thus n = 100 is the minimum required sample size.
5.53. Random samples from the entire population of interest reduce the chance of a bias and increase chance of being representative of the entire population. Also, we have a known probability of being within certain distances of the parameter of interest. We use a frame and a random number
generator or a table of random numbers. A simple random sample is such that every possible set of n elements has an equal chance of being selected.
5.54. A bias is a systematic deviation away from the target of estimation. A bias takes us away from the target parameter in repeated sampling. If the bias is small and variance of the estimator is also small, the bias may be tolerated, especially if the bias decreases as n increases.
5.55. The sample median is unbiased. The sample mean is more efficient; it is also sufficient. This is why we prefer the sample mean. We must assume normality for using the sample median to estimate
. The median is more resistant to outliers.5.56. S 2 has n – 1 in the denominator because there are n – 1 degrees of freedom for deviations from the sample mean. Using n – 1 instead of n makes S 2 an unbiased estimator of 2.
5.57.
= 44
= 7 n = 50P(X < 35) = P(Z < -9.0918) = .5 .5 = 0.00 Sampling Distribution of Sample Mean
known
Population Distribution
Mean Stdev Is the population normal?
44 7
Sample Size Sampling Distribution of X-bar
n 50 Mean Stdev
44 0.98995
P(<x) x
Symmetric Intervals x1 P(x1<X<x2) x2 18.46122 0.95 20.538779 18.62823 0.9 20.371772 5-59.
= 3.9
= 0.5 n = 25 P(X
> 3.0) = P 25 / 5 . 0 9 . 3 0 . 3 Z = 0.5 + 0.5 = 1.000 (Use template: Sampling Distribution.xls, sheet: x-bar) Sampling Distribution of Sample MeanPopulation Distribution
Mean Stdev
3.9 0.5
Sample Size Sampling Distribution of X-bar
n 25 Mean Stdev 3.9 0.1 x P(>x) 3 1.0000 5.60.df = (rows-1)(columns-1) = (5-1)(3-1) = 8 5-61. p = 0.38 n = 100 P(Pˆ > 0.30) = P 100 / ) 62 )(. 38 (. 38 . 30 . Z = P(Z > 1.648) = .5 + .4503 = 0.9503 where stdev = SQRT(.38*.62)
Sampling Distribution of Sample Mean
Population Distribution
x P(X>x)
0.3 0.9503
5-62. X is normal. But since
is unknown and we use S, the quantity (X
)/(S/ n ) has the t(n1) distribution rather than the standard normal distribution Z.5-63. No minimum (n = 1 is enough for normality).
5.64. X , Pˆ , S 2 are unbiased. S is the square root of an unbiased estimator of 2, thus it is not unbiased. Proof:
Assume E(S) =
then: (E(S))2 = 2and: E(S 2) – (E(S))2 = 2 2 = 0 (since E(S 2) = 2). But E(S 2) – (E(S))2 = V(S)
V(S) = 0 means that S is not a statistical estimator. The contradiction establishes the proposition that S is biased.
5.65. This estimator is also consistent. It is more efficient than X , because 2/n 2 < 2/n. 5.66. df = 124 –3 = 121
a. Normal population requires the smallest minimum n.
b. Mound-shaped population requires the next higher minimum n. c. Discrete population needs the highest minimum n.
d. Slightly skewed population: n more than for (b), less than for (c). e. Highly skewed population: n less than for (c), but more than for (d). The relative minimum required sample sizes are as follows:
a
n < n < b n < d n < e nc
5.67. Yes. SE(X ) decreases as n increases:
SE(X ) =
/ n , which goes to 0 as n goes to
. Statistically, it is always good to have as5.69. P(Pˆ < .15) = P 250 / ) 8 )(. 2 (. 20 . 15 . Z = P(Z < 1.976) = .5 .4759 = 0.0241 5-71.
= 25
= 2 n = 100 P(X < 24) = P 10 / 2 25 24 Z = P(Z < 5) = 0.0000003 Not probable at all.5.72.P
Pˆ0.60 0.07
= P(0.53 Pˆ 0.67) = P 200 / ) 40 )(. 60 (. 60 . 67 . 200 / ) 40 )(. 60 (. 60 . 53 . Z = P(2.02
Z
2.02) = 0.9567Sampling Distribution of Sample Proportion
Population
Proportion
p
0.6
Sample Size Sampling Distribution of P-hat
n 200 Mean Stdev 0.6 0.03464 x1 P(x1<P hat<x2) x2 0.53 0.9567 0.67 5.73.P(1.52 < X < 1.62) = P 200 / 4 . 0 57 . 1 62 . 1 200 / 4 . 0 57 . 1 52 . 1 Z = 2TA(1.768) = 0.923
5-74.
a) point estimate for the sample mean is 52
Population Distribution
Mean Stdev Is the populationnormal?
52 2.4
Sample Size Sampling Distribution of X-bar
n 40 Mean Stdev
52 0.37947
P(X<x) x x P(X>x) x1 P(x1<X<x2) x2
52 0.4958 53
b) P( 52 < X < 53) = 0.4958
5-75 (Use template: Sampling Distribution.xls, sheet: p-hat) n = 400 p = 0.06
Sampling Distribution of Sample Proportion
Population Proportion
p
0.06
Sample Size Sampling Distribution of P-hat
n 400 Mean Stdev
0.06 0.01187
P(P-hat < 0.05) = 0.1999
P(<x) x
0.1999 0.05
Sample Size Sampling Distribution of X-bar n 10 Mean Stdev 15830 144.832 1202 . 0 ) 16000 X ( P x P(>x) 16000 0.1202
5-77 (Use template: Sampling Distribution.xls, sheet: x-bar) μ = 3.42 σ = 1.5 n = 30
Sampling Distribution of Sample Mean
Population Distribution
Mean Stdev
3.42 1.5
Sample Size Sampling Distribution of X-bar
n 30 Mean Stdev 3.42 0.27386
0171
.0
)
00
.4
(
X
P
x P(>x) 4 0.01711) 0.6210. No it is not an acceptable level of performance
Mean Stdev Is the populationnormal?
1.008 0.045
Sampling Distribution of X-bar
50 Mean Stdev
1.008 0.00636
P(X<x) x x P(X>x) x1 P(x1<X<x2) x2
0.99 0.6210 1.01
2) 1.00 This would result in the lowest SSD and MSD 3) 0.01104
Population Distribution
Mean Stdev Is the populationnormal?
1.008 0.01104
Sample Size Sampling Distribution of X-bar
n 50 Mean Stdev
1.008 0.00156
P(X<x) x x P(X>x) x1 P(x1<X<x2) x2
6)
Reduction P(0.99 < X <
1.01) Stdev stdev Cost
0.9 0.01104 33.96 $ 172,992.24 0.95 0.00861 36.39 $ 198,634.82 0.99 0.00609 38.91 $ 227,098.22 7) Reduction Total P(0.99 < X <
1.01) Stdev stdev Cost Cost
0.9 0.04302 1.98 $ 588.06 668.06 $ 0.95 0.03609 8.91 $ 11,908.22 $ 11,988.22 0.99 0.02749 17.51 $ 45,990.02 $ 46,070.02 8) Use a mean of 1.00 and adjust the standard deviation to 0.02749