Example 7.8 – Application of Concepts
a.
Where the mean strength of the new concrete block surface is no different (meaning equal)
from that of the original surface, this condition precisely describes the properties of the
sampling distribution of the sample mean, where the sampling distribution of the sample mean
is equal to the mean of the population mean from which it was selected. The sampling
distribution is equal to the mean of the sampling population.
b.
Find the probability that sample mean and the sample mean load classification (LCN) of the 25
concrete block sections, exceeds 65. The sampling distribution is approximately normal
Step 1: Compute the desired area by obtaining the z score for x̄ =65
Answer:
Z = (x ̄- 𝜇𝑋 )/σX
=
65−60
2
= 2.50
Step 2: Construct a probability equation based upon the question.
Answer:
P (x̄ ≥ 65) = P (z ≥ 2.50)
= .5 – A (source; Figure 7.12)
= .5 - .4938
= .0062
c.
What can be inferred about the true mean LCN of the new surface?
Answer:
The original assumption about the concrete blocks was incorrect.
Step-by-step generalized solution summary for examples: 7.8, 8.4 – 8.8, 8.12 – 8.15, 8.19, 8.21, and 8.30
1
Example 8.4 – Theoretical Interpretation of a Confidence Level
Question:
Interpret the result for the 95% Confidence Intervals for µ for 40 Random Samples of
100 sale Prices from Appendix A.1.
Answer:
Since the majority of the 40 repetitions of the confidence interval procedure described
contain the mean of the population ($106,405), we can be assured that the statistical
procedure used will result in and value that represents the same majority of the
population.
Example 8.5 – Finding Z for a 90% Confidence Interval
Question:
Determine the value of 𝑍∝/2 that would be used in constructing a 90% confidence
interval of a population mean based on a large sample.
SOLUTION
Step 1:
Use the given value of α/2 = .05 to confirm that an area of .05 in the upper tail f the
standard normal distribution and since the total area to the right of 0 is .50, z.05 is such
that the area between 0 and z.05 is .050 - .05 = .45.
Step 2:
Find the corresponding table, find z.05 = 1.645
We know that the sample mean is ± 1.645𝜎 subscript the sample mean
Example 8.6 – 99% Confidence Interval
Question:
Estimate µ, the mean ration of sale price to total appraised value for all properties sold
in 1993, using a 99% confidence interval. Interpret the interval in terms of the problem.
SOLUTION
Step 1:
Establish the sample mean
2
𝜎
√𝑛
Answer
x̄ ± 2.58
Step 2:
Substitute the values for the sample mean and standard deviation
Answer:
1.315 ± 2.58 (
.366
√50
)
With a confidence level of 99% that the interval encloses the true mean ratio of sale
price to total appraised value for all residential properties sold in six Tampa
neighborhoods in 93’. We can also conclude that there is a general tendency for the sale
prince of a property in these neighborhoods to exceed its total appraised value. This
variance is likely the result of human error in the process of appraising.
Example 8.7 – Effect of (1-α) on the Width of the Confidence Interval
Question:
a.
Construct a 95% confidence interval for the mean ratio of sale price to total
appraised value for properties sold in the six neighborhoods.
SOLUTION
Step 1:
Answer:
x̄ ± 1.96
𝜎
√𝑛
≈ x̄ ± 1.96
𝑠
√𝑛
= 1.315 ± 1.96 (
.366
√50
)
= 1.315 ± .101 or (1.214, 1.416)
Question:
b. For a fixed sample size, how is the width of the confidence interval related to the
confidence coefficient?
SOLUTION
Answer:
The Relationship between width of confidence interval and confidence coefficient
guideline informs us that as the width increases we have a greater chance that the
width will contain the true parameter value.
Example 8.8 – Effect of n of the Width of the Confidence Interval
Question:
a.
Construct a 99% confidence interval for µ, the population mean ratio of sale
price to total appraised value.
SOLUTION
3
Step 1:
Substitute the values of the sample statistics into the general formula for a 99%
confidence interval
Answer:
x̄ ± 2.58
𝜎
√𝑛
≈ 1.315 ± 2.58 (
.366
√50
)
= 1.315 ± .094
Question:
b. For a fixed confidence, how is the width of the confidence interval related to the
sample size?
Answer:
The relationship between width and confidence interval and sample size guideline
informs us that the width of the confidence interval decreases as the sample size
increases. Simply, larger samples provide more information than do small samples.
Example 8.12 – Computer Analysis
Question:
Referring to example 8.6, compare the results to the interval calculated using the t score
and z scores.
Answer:
The computed interval calculated by the MINITAB uses a t score when the population
standard deviation is unknown. A calculated z score renders a less precise interval
because in this case the population parameters are not well defined. This is precisely
why computers use a t score.
Example 8.13 – Selecting a Point Estimate
Question:
Estimate 𝜋, the true proportion of all state lottery winners (at least $50,000) who quit
their jobs during the first year after striking it rich.
Step 1:
Calculate the sample proportion p
Answer:
p=𝑛=
𝑥
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑙𝑜𝑡𝑡𝑒𝑟𝑦 𝑤𝑖𝑛𝑛𝑒𝑟𝑠 𝑖𝑛 𝑠𝑎𝑚𝑝𝑙𝑒 𝑤ℎ𝑜 𝑞𝑢𝑖𝑡 𝑗𝑜𝑏
𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑙𝑜𝑡𝑡𝑒𝑟𝑦 𝑤𝑖𝑛𝑛𝑒𝑟𝑠 𝑖𝑛 𝑠𝑎𝑚𝑝𝑙𝑒
63
P = 576 = .11 (11%)
4
Example 8.14 – 95% Confidence Interval for 𝝅
Question:
Construct a 95% confidence interval for𝜋, the population proportion of state lottery
winners who quit their job within 1 year of striking it rich.
Step 1:
Substitute the values into the formula for the 95% confidence interval
Answer:
𝜌 ± 𝑧α/2 √ 𝑛 = .11 ± 1.96 √
𝑝𝑞
(.11)(.89)
576
= .11 ± .03
Example 8.15 – 90% Confidence Interval for 𝝅
Question:
Estimate 𝜋, using a 90% confidence interval. Interpret the interval
SOLUTION
Step 1:
Define the sample proportion of families that watched the premiere of “Seinfeld”
Answer:
p=𝑛=
𝑥
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑎𝑚𝑖𝑙𝑖𝑒𝑠 𝑖𝑛 𝑠𝑎𝑚𝑝𝑙𝑒 𝑡ℎ𝑎𝑡 𝑤𝑎𝑡𝑐ℎ𝑒𝑑 𝑡ℎ𝑒 𝑝𝑟𝑒𝑚𝑖𝑒𝑟𝑒
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑎𝑚𝑖𝑙𝑖𝑒𝑠 𝑖𝑛 𝑠𝑎𝑚𝑝𝑙𝑒
101
P = 165 = .612
Step 2:
Insert q into the 90% confidence formula
Answer:
p ± z.05 √
𝑝𝑞
𝑛
(.612)(.388)
= .612 ± 1.645 √
165
= .612 ± .062
Example 8.19 – Small Sample 95% Confidence Interval for µ1 - µ2
Questions:
a. Construct a 95% confidence interval for the difference between the mean buyer
savings of the two strategies.
SOLUTION
Step 1:
Let µ1 and µ2 represent the true mean savings of buyers using the competitive and
comparative bargaining strategies.
Step 2:
Compute an estimate of this common variance
5
Answer:
𝑆𝑝2 =
=
(𝑛1−1)𝑆12 +(𝑛2−1)𝑆22
𝑛1+𝑛2−2
(8−1)(538)²+(8−1)( 357)²
8+8−2
= 208,446
Step 3:
Substitute the appropriate quantities into the general formula and solve
Answer:
(1,706 – 2,106) ± 2.145 √208.446 (8 +
1
1
)
8
= -400 ± 490
Example 8.21- 95% Confidence Interval for µd
Question:
Find a 95% confidence interval for the difference in mean levels of assertiveness, µd =
(µ1 - µ2).
SOLUTION
Step 1
Make necessary assumptions based on the sample size.
Answer:
Since the sample is small (n = 10) we must assume that the differences are from an
approximately normal population.
Step 2:
Substitute the values
=11.0 ± 4.7
Example 8.30 – 95% Confidence Interval for 𝝈²
Question:
Construct a 95% confidence interval for the true variation in tar contents of domestic
cigarette brands.
SOLUTION
Answer:
(𝑛−1) 𝑠²
𝑋²𝑎/2
≤ 𝜎² ≤
(𝑛−1)𝑠²
𝑋2 𝛼
(1− 2 )
Step 1: Substitute values
6
Answer:
(500−1)(4.93)²
563.852
≤ 𝜎² ≤
(500−1)(4.93)²
439.936
21.51 ≤ 𝜎² ≤ 27.57
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