HW #5
Example 7.8
SOLUTION STEPS:
a. Determine:
i. population and sample means
ii. population and sample standard deviations.
b. Review x data to understand how the number was derived
c. The problem is normally distributed so central limit theory can used
b. The sampling distribution normally distributed
a. probability of P( x ≥ 65) can determine using the z score where
i. Z = 2.50
ii. P( x ≥ 65) = P(z ≥ 2.50)
1. = .5 - .4938 (the area between 60 – 65 under a normal distribution)
2. = .0062 (the area under the curve beyond the mean of 65)
c. Conclusion
a. Observing anything beyond 65 is highly unlikely
i. x = 73 can be classified as a rare event if µ = 60.
Example 8.4
SOLUTION STEPS:
 Identify the given data
o 40 random samples of 100 sale prices
 µ =$106,405
 95% confidence interval
 Results confirm that 95% of the 40 samples do contain the µ value and should be within
the upper/lower limits.
Computer Implementation: n/a
Example 8.5
SOLUTION STEPS:




Determine type of distribution:
o normal distribution contains a mean of zero
A 90% confidence interval is subtracted from the total probability (100%)
o 10% or 0.10
 0.10 is divided by 2 to get .05.
Find z.05.
o half of the normal table we subtract.5-.05 = .45.
Identify the value in the standard normal table for .45
o 1.64 corresponds to .4495 and 1.65 corresponding to .4505
1


Interpolating between the values
 The corresponding Z =1.645.
Conclude the confidence interval is x =1.645 ±
Example 8.6
SOLUTION STEPS:




Determine the mean and standard deviation for the sample using Excel
o
=1.315 and s=.3658
Confidence coefficient for 99% = 2.58
Use the sample standard deviation, s
o s is sufficiently large: 50
Identify upper and lower limits
o Upper limit

=1.3158 ± 2.58(
) = 1.449
o Lower limit
=1.315 ± 2.58(
) = 1.18
We are 99% confident that the true mean ratio of sale price is contained within the
interval from 1.18 to 1.45
Computer Implementation:
Using the data analysis add-in package in Excel I was able to identify the values and upper and
lower limits stated above.
Example 8.7
SOLUTION STEPS:


construct a 95% confidence interval
o confidence coefficient given:
 z = 1.96 and x = 1.315 ± 1.96 (
) = 1.315±.101.
 lower limit of 1.214 and upper limit of 1.416
o It can be determined the C.I. has widened or increased with the decrease in
confidence from 99% to 95%.
Due to the issue stated above the confidence interval coefficient must also decrease
x = 1.315 ± 1.96(
) 95% Confidence
x = 1.315 ± 2.58(
) 99% Confidence
Computer Implementation: n/a
Example 8.8
SOLUTION STEPS:
2

Determine the effect of a change in sample size changes from n=50 to n=100
o 99% Confidence Sample size n=50
 Upper limit 1.449 / lower limit 1.18

x = 1.315 ± 2.58(
)
x = 1.315 ± .1335
o 99% Confidence Sample size n=100
 Upper limit 1.409 / lower limit 1.221


x = 1.315 ± 2.58(
)
x = 1.315 ± .0944
The about results indicate the width of the confidence interval increases as sample size
decreases
Example 8.12
SOLUTION STEPS:
 The confidence interval from the computer program has a lower limit of 1.1766 and
upper limit of 1.4538.
o The confidence interval from my calculation was lower limit of 1.18 and, upper
limit of 1.45.
 Actual data 1.182258 and 1.449343
 Standardize how far we carry the decimal point then the results should be
relative the same
 The computer program apparently uses the t statistic to calculate the
confidence interval
Computer Implementation:
Excel mimics the same program used above with some slight variations.
Example 8.13
SOLUTION STEPS:
 Identify equation parts:
o Sample size is n= 576 respondents
o 63 quit their jobs within 1 year after winning $50,000
 p = the proportion of the sample
 x = the number of respondents that quit their job
 n = sample population
 Divide that part of the population over the total sample population
o = 63/576
 = .11 or 11%
3
o 11% of the lottery winners in the sample quit their job during the first year after
winning
Computer Implementation:
Excel mimics the same program used above with some slight variations.
Example 8.14
SOLUTION STEPS:


construct a 95% confidence interval using the formula: p±
o p=.11.
o Coefficient is 1.96 (95% C.I.)
o q=1-p
o n=576
o The 95% C.I. is: 0.08, 0.14 where 0.11± = 0.11± 0.03
Computer Implementation: N/A
Example 8.15
SOLUTION STEPS:
 Find the random sample of 165 families
o 101 responded positively to the question asked. p=
= 101/165 = .612.

Use a 90% C.I. using p±

.612±1.645
= .612±.062.
o A 90% confidence interval of 0.55, 0.674)
We can conclude we are 90% confident that the true proportion of TV families that
watched the show is between .550 and .674

=
Example 8.19
SOLUTION STEPS:
 Identify the given parts
o Two type of bargaining strategies - competitive (
versus coordinative (
o Each sample n=8.
 Construct a 95% confidence interval for the difference between the buyer savings of the
two strategies.
4
o Assume :
1. The population of each have approximate normal distributions
2. The variances of the two are equal
3. The samples were random and unbiased.


Determine variance
o adding the two together and dividing by 2.
 Substituting our values we get (-400±490) or (-890, 90)
Result: with a 95% confidence estimate that the difference (
) falls within the interval
-890 to 90
Computer Implementation:
The above calculations were performed in Excel (see below) using the Excel functions for
average (AVG), standard deviation (STDEV), and variance (VAR)
N
Competitive
Coordinative
1
1857
1544
2
1700
2640
3
1829
1645
4
2644
2275
5
1566
2137
6
663
2327
7
1712
2152
8
1679
2130
SUM
13650
16850
x
1706.25
2106.25
s
538
357
289,432
127,797
208,614
AVG variance
Example 8.21
SOLUTION STEPS:
 Identify given that data and information
o Find the 95% confidence interval for the difference in assertiveness for the two
different training methods
o Find the avg and s
 Avg 11.0, and s of 6.53.
o Add all values to the equation
 11.0 ±4.7 or (6.3, 15.7).
 Result with 95% confidence that the mean managerial assertiveness test scores fall within
the interval 6.3 to 15.7
o Both interval limits are positive
5
 method 1 produces a test score that is larger
o In general the test scores for method 1 are higher
Computer Implementation:
Using Excel I was able support the average and s stated in the problem.
Example 8.30
SOLUTION STEPS:






Identify given data
o x = 11.06 mg and s = 4.93 mg for 500 cigarette brands.
Construct a 95% confidence interval for .
of .025
Determine which formula to use
o
.
Find the value on the table for chi squire
o x2 .025 and x2 .975 with n=500 = degrees of freedom.
 values are 21.51 ≤ σ2 ≤ 27.57.
Result with 95% confidence that the true variance falls between 21.51 and 27.57
Computer Implementation:
By using the descriptive statistics function in Excel the mean, variance, and standard deviation
were generated among other things. However, there was a difference between the average
variance in Excel and the book, but the difference was mainly where the decimal point was
placed. When the square root was taken for each number to determine the standard deviation the
difference was decreased even further.
6