CHAPTER 9
ESTIMATION AND CONFIDENCE INTERVALS
1. See definition on page 362 of the text.
3. See definition on page 363 of the text.
5.(a)Standard error of the mean = ; (1 - ) = (1 – 0.1) = 0.9. Since the population distribution is approximately normal, the distribution of is approximately normal. Hence, margin of error = ; 90% confidence interval =
(b)Standard error of the mean = ; (1 - ) = (1 – 0.025) = 0.975. Since sample size is large enough (n = 50 > 30), the distribution of is approximately normal. Hence, margin of error = ; 97.5% confidence interval =
(c)Standard error of the mean = ; (1 - ) = (1 – 0.01) = 0.99. Since sample size is large enough (n = 60), the distribution of is approximately normal. Hence, margin of error = ; 99% confidence interval =
7. Since the population is approximately normally distributed, the distribution of is approximately normal. Hence, a 90 percent confidence interval estimate for is
=
=
9.(a)It is the number in the table of Student’s t Distribution in row corresponding to “18” and column corresponding to “” . It equals 34.8053.
(b)Since the required left tail probability is 0.025, the right tail probability is 1 – 0.025 = 0.955. It is the number in the table of Student’s t Distribution in row corresponding to “30” and column corresponding to “”. So, L = 16.7908.
(c)Total area to the right of is 0.99. We want area between and U to be 0.98. Hence, area to the right of U should be (0.99 – 0.98) = 0.01. Thus, . It is the number in the table of Student’s t Distribution in row “5” and column “” and equals 15.0863.
(d)We want area between L and to be 0.925. Area to the left of is 0.05. Hence, total area to the right of L should be 0.925 + 0.05 = 0.975. Thus, . It is the number in the table of Student’s t Distribution in row “15” and column “” and equals 6.26214.
11.(a)Since the population is normal, follows chi-square distribution with df = (n – 1) 9.
A 99 percent confidence interval estimate for 2 is
(b)An 80 percent confidence interval estimate for 2 is
13.(a)(1 - ) = 0.8, so = (1 – 0.8) = 0.2.
Hence, t = t/2 = t0.1 (for df = 30) = 1.31 (from the t-table)
(b)From the t-table, we find that t0.005 (for df = 6) = 3.707.
(c)By symmetry of t-distribution it follows that t = -t0.1.
From the t-table, we get t0.1 (for df = 24) = 1.318. Hence, t = -1.318.
(d)The right tail area corresponding to t = 1 – 0.9 = 0.1. Thus, t = t0.1 (for df = 12) = 1.356 (from the t-table).
(e)The left tail area corresponding to t0.995 is (1 – 0.995) = 0.005. Thus, t0.995 = -t0.005.
From the t-table, we get t0.005 (for df = 14) = 2.977.
Hence, t0.995 = -2.977.
15.(a)Since the population distribution is approximately normal, the distribution of is approximately t-distribution with df = n – 1.
Calculate and s. (1 - ) = 0.95. So, = (1 – 0.95) = 0.05. For df, t/2 = t0.025 = 2.201. 95% confidence interval estimate for is .
(b)Calculate and s. (1 - ) = 0.9. So, = (1 – 0.9) = 0.1. For df = 19, t/2 = t0.05 = 1.729. 90% confidence interval estimate for is .
(c)Calculate and s. (1 - ) = 0.99. So, = (1 – 0.99) = 0.01. For df = 7, t/2 = t0.005 = 3.499. 99% confidence interval estimate for is .
17.
Since the population distribution is approximately normal, the distribution of is approximately t-distribution with df = n – 1 = 19.
(1 - ) = 0.99. So, = (1 – 0.99) = 0.01. For df = 19, t/2 = t0.005 = 2.861
99% confidence interval estimate for is
The value 50 lies in the 99% confidence interval estimate. Hence, the analysis does not provide reason to rule out the probability that the value of = 50. The value 60 does not lie in the interval estimate. Hence it will be unreasonable to assume that = 60.
19.(a) The best point estimate of is the sample mean= 21.9 eggs/month.
(b) (1 - ) = 0.98. So, = (1 – 0.98) = 0.02. For df = 19, t/2 = t0.01 = 2.539
Hence, a 98% confidence interval estimate for is
=
= = (20.708,23.092)
(c) The population is normally distributed, but population standard deviation is unknown. In this case has t-distribution with (n - 1) degrees of freedom.
(d)Value 22 lies in the 98% confidence interval estimate. Hence, the analysis does not provide evidence against the probability that the value of = 22. Value 24 does not lies within confidence interval estimate. Hence it will be reasonable to assume that the value of is not 24.
21.(a) The best estimate of the value of the population proportion is .
(b) An estimate of the standard error of the proportion is
(c) (400)(0.75) > 5 and (400)(0.25) > 5. Hence, we can assume normal approximation. An approximate 99% confidence interval estimate of population proportion is
=
= .
(d)The estimator used in (c) for finding the confidence interval estimate produces an interval containing the population proportion, p, 99 percent times in the long run. Given the high probability, it would be reasonable to expect the interval currently obtained to contain the value of p.
23.(a) The best point estimate of the population proportion is .
(b) (300)(0.05) > 5 and (300)(0.95) > 5. Hence, an approximate 95% confidence interval estimate of p is
=
=
(c)The entire 95% confidence interval estimate obtained in (b) lies to the left of 0.1. Hence, it is reasonable to infer that p is less than 0.1. He should not return the lot.
25. Let us hope that the sample size, n, will be large enough for the distribution of to be approximately normal. Then, the sample size n should be at least = or 60.
This value of n is large enough for assumption of normality. So the bound is valid.
27.Let us hope that the sample size, n, will be large enough for normal approximation.
We want (1 - ) = 0.99, so = 0.01. z = z0.005 = 2.575.
An approximate lower bound on the required sample size is or 165.
This value of n is large enough for assumption of normality. So the bound is valid.
29. We want the margin of error to be no more than + 0.05. Let us hope that the sample size will be large enough for distribution of to be approximately normal.
An approximate lower bound on the required sample size is
= or 62.
This value of n is large enough for assumption of normality. So the bound is valid.
31.(a)Let us hope that the sample size will be large enough for applying normal approximation.
An approximate lower bound on the required sample size is =or 5683.
This value of n is large enough for assumption of normality. So the bound is valid.
(b)One way is to first obtain a better estimate of p by first choosing a pilot sample of moderate size, say 100. If the sample size based on new estimate is still large, then we shall have to increase the allowable margin of error and/or decrease the confidence level. For example, if the allowable margin of error is changed to 0.05, then an approximate lower bound on sample size n will be = 227.3 or 228.
33. The ratio (n/N) = (49/500) = 0.098 is greater than 0.05, but not too large.
The population size is moderately large, and the population distribution is approximately normal. Hence, we shall use Formula 9-18 in the textbook.
(1 - ) = 0.99. So = (1 – 0.99) = 0.01. t/2 = t0.005 (for df = 48) is approximately 2.68.
An approximate 99% confidence interval estimate for is:
= =
= (36.724, 43.276).
35.The ratio (n/N) = (30/300) = 0.1 is greater than 0.05, but not too large.
.
=30 (0.6) > 5 and = 30 (40) > 5.
Hence n is large enough for normality assumption.
An approximate 95% confidence interval estimate for p is:
=
= (0.433, 0.767).
37. Let us hope that the sample size will be large enough for assumption of normality.
An approximate lower bound for the sample size is
= , or 2185.
This value of n is large enough for normality assumption. So the bound is valid.
39.(a)The best point estimate 54. That is, the best choice of a single value as an estimate of is 54.
(b)n = 49 is large enough for assumption of normality of distribution of . For df = 48, t0.025 = approximately 2.01. A 95% confidence interval estimate is.
41. Let us hope that the sample size will be large enough for assumption of normality.
An approximate lower bound on the required sample size is = or 97.
This value of n is large enough for assumption of normality. Hence the bound is valid.
43.(a) A point estimate of population mean is 1.01 kg.
(b)The sample size (n = 36) is large enough for us to assume that distribution of is approximately student’s t-distribution
For df = 35, t0.025 = approximately 2.031. A 95% confidence interval estimate is.
45.(a)The sample size (n = 50) is large enough for assumption of normality.
For df = 49, t0.005 = approximately 2.68.
A 99% confidence interval estimate is = .
(b)The result, thus provides significant evidence that value of is not 0.63. It will therefore be reasonable to conclude that the value of is not 0.63.
47.(a)The value of sample proportion is = 0.63.
n = (1000) (0.63) > 5 and n (1 - ) = (1000) (0.37) > 5.
Hence, the sample size is large enough for normality assumption.
A 95% confidence interval estimate is
= .
(b) The entire interval in (a) lies above 0.6. The analysis thus provides sufficient evidence that the population proportion p is greater than or equal 0.6. It is therefore reasonable to conclude that the claim, that p is less than 0.6 is false.
49. The sample size is large enough to assume normality.
For df = 35, t0.05 = approximately 1.69. A 90% confidence interval estimate of is = .
51.(a)Let us hope that the sample size will be large enough for assumption of normality. = 0.21.
An approximate lower bound for n is = or 709.
This is large enough for assumption of normality. Hence the bound is valid.
(b)An approximate lower bound on the required sample size n is = or 1068.
Hence, 1068 accountants should be contacted.
53. Let us hope that the sample size will be large enough for assumption of normality. = 5/50 =0.1.
An approximate lower bound for n is =
An approximate lower bound for n is or 865. This is large enough for assumption of normality. Hence the bound is valid.
55.(a).
Thus the best point estimate of is 89.4667.
(b)The sample mean is
Since the population distribution is approximately normal, the distribution of is approximately t-distribution with df = n – 1 = 14.
For df = 14, t0.025 = 2.145. Hence, 95% confidence interval estimate is .
(c)The confidence interval estimate in (b) lies entirely above 80. Hence it is reasonable to conclude that the mean stress level is in the dangerous level.
57. The sample size (n = 60) is large enough for assumption of normality.
= (1 – 0.98) = 0.02. For df = 59, t/2 = t0.01 = approximately 2.39. Hence, an approximate 98% confidence interval estimate of is .
59.(a) The best point estimate of the population mean, , we can obtain is .
(b)Since the population distribution is approximately normal, the distribution of is approximately t-distribution with df = n – 1 = 9.
The sample standard deviation is
= 304.4276.
For df = 9, t0.025 = 2.262.
A 95% confidence interval estimate for is .
61.(a)
(b)Since the population distribution is approximately normal, the distribution of is approximately t-distribution with df = n – 1 = 11.
The sample standard deviation is
For df = 11, t0.05 =1.796. Hence, 90% confidence interval estimate is.
(c)The value 60 lies in the interval estimate obtained in part (b). Thus, our analysis does not provide significant evidence against the claim. We have no reason to doubt the claim.
63.A point estimate of proportion, p, of the entire population (Ontario) who blame the provincial government for the tragedy is = 0.22.
n = (1001) (0.22) > 5 and n (1 - ) = (1001) (0.78) > 5.
Hence, the sample size is large enough for normality assumption.
An approximate 95% confidence interval estimate for p is
= = (0.194, 0.246).
65.(a)A point estimate of proportion, p, of Canadians who favour user fees and charges is = 0.54.
n = (1400) (0.54) > 5 and n (1 - ) = (1400) (0.46) > 5.
Hence, the sample size is large enough for normality assumption.
An approximate 95% confidence interval estimate is = .
(b)The entire interval lies above 0.5. Hence, it is reasonable to conclude that p is greater than 0.5, that is, majority of Canadians favour moderate user fees and charges.
67. Select a sample of size 10; calculate and s. For df = 9, t0.025 = 2.262. A 95% confidence interval estimate is . The answer will vary according to the sample chosen.
69.The Minitab and Excel outputs are given below. (See the Minitab and Excel instructions given in the chapter.)
It should be noted that Minitab output directly gives the confidence interval estimates, while the Excel output gives in the row labeled “Confidence level”, the margin of error.
(a)Using the Excel output, we get for selling price of the homes:
95 percent confidence interval estimate for population mean
=
= = (211.9868122, 230.2189021).
(b)A 90 percent confidence interval estimate for the fraction of home that were sold more than $240000 = .
(c)For number of bedrooms: 90 percent confidence interval estimate for the population mean =
= = (3.45354, 4.14646).
(d)A 90 percent confidence interval estimate for the proportion of homes sold that have a garage = = .
(e)A 95 percent confidence interval estimate for the proportion of homes sold that have two or more bathrooms = (0.779, 0.916)
71. The Minitab and Excel outputs are given below. (See the Minitab and Excel instructions given in the chapter.)
It should be noted that Minitab output directly gives the confidence interval estimates, while the Excel output gives in the row labeled “Confidence level”, the margin of error.
Thus, using Excel, we get :
For TSE 300 data : 92 percent confidence interval =
For BCE data : 95 percent confidence interval = .
For Air Canada data : 90 percent confidence interval =
These are the same (except for rounding) as the ones obtained using Minitab.
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