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I. Sampling Distribution Problems
1. #7.19, p. 292
Preliminary:
Let X denote the appraised value of a home on the Southwest side
of Denver. The average value of all homes (N = 1500 houses) in the Southside is µ
= $137,000 with a population standard deviation of Ó =$8500. Suppose n
=100 homes are randomly selected from this neighborhood. Find the probability that
x-bar value from a sample of 100 will be greater than $145,000?.
Notes:
i) Because n/N is greater than .05, the finite population correction factor [N-n/N-1]½
must be used in order to enhance the precision of the X-bar via its standard
error (Óx-bar).
ii) According to the CLT, X-bar is distributed as normal with E(X-bar)
= µ and the standard error of X-bar is given as Óx-bar
= Ó/n½ provided "n" is statistically large.
So Óx-bar = Ó/n½ [N-n/N-1]½ . The
result of the calculation is Óx-bar = 8500/100½
[.9664] = $821.44.
P( X-bar > 145,000) implies P( Zi > ZLL) where ZLL = (X-bar - µ)/Óx-bar = (145,000 - 137,000)/821.44 = 9.739 or 9.4. Thus P(X-bar > 145,000) = P(Zi > 9.5) = .5000 - .5000 = .0000. This means that it is impossible that a sample of 100 houses will produce a sample mean value that is greater than $145,000
2. #7.29, p. 296
Preliminary:
Let X denote the # of customers that purchase Motorola's cell phones. Because Motorola
commands 32% of the cell phone market it means that P = .32.
Let Y denote the # of customers that purchase some brand other than Motorola's cell
phones. Because Motorola commands 32% of the cell phone market it means that
other brands command 68% of the market in which case P = .68 is the
proportion of all non-Motorola cell phones sold by the industry.
Suppose n = 200 recent cell phone buyers are randomly selected.
a) What is the probability that fewer than 50 purchased Motorola cell
phones?
X = 50 implies that P-hat = 50/200 = .25. Thus P X < 50) = P(P-hat
< .25).
According to the CLT in repeated sampling P-hat has a normal
distribution with E(P-hat) = P, and its standard error (Óp-bar
) is given as Óp-bar = [P(1-P)/n]½ = [(.32 x .68)/200]½
= .0329
Thus P(P-hat < .25) implies P(Zi < ZUL)
where ZUL = (P-hat - P)/Óp-bar
= (.25 - .32)/.0329 = -2.1276
So that P(Zi < -2.13) = .5000 - .4834 = .0166. This means it is
highly unlikely that of the 200 recent cell phone buyers fewer than 50 bought a
Motorola brand.
b) What is the probability that more than 70 purchased a Motorola brand?
X = 70 implies that P-hat = 70/200 = .35. Thus P X > 70) = P(P-hat
> .35).
P(P-hat > .35) implies P(Zi > ZLL) where
ZLL = (P-hat - P)/Óp-bar =
(.35 - .32)/.0329 = .9118
So that P(Zi > .9118) = .5000 - .3186 = .1814. This means there is
an 18% chance that of the 200 recent buyers of cell phones more than 70 of them bought
a Motorola brand.
c) What is the probability that less than 140 purchased some brand other than
Motorola?
Y = 140 implies that P-hat = 140/200 = .70. Thus P (X < 140) = P(P-hat
< .70).
P(P-hat < .70) implies P(Zi < ZUL) where
ZUL = (P-hat - P)/Óp-bar =
(.70 - .68)/.0329 = .61
So that P(Zi < .61) = .5000 + .2291 = .7291. This means there is
about 73% chance that of the 200 recent buyers of cell phones less than 140 of
them bought some brand other than Motorola.
II. Interval Estimation Problems
1. The duration (in minutes ) for a sample of 20 flights-reservation
telephone calls is shown in as follows:
| 2.1 | 10.4 | 4.8 | 5.5 | 5.9 | 10.5 | 4.5 | 4.8 | 3.3 | 5.8 |
| 2.8 | 6.6 | 7.5 | 4.8 | 5.5 | 3.5 | 5.3 | 3.6 | 7.8 | 6.0 |
Preliminary:
Let X
denote the duration of
flight-reservation phone calls (in minutes).
n=20 implies that X-bar = 5.55 minutes and S = 2.22 where S is the unbiased
sample standard deviation.
a) What is the point estimate of the population mean time for
flight-reservation telephone calls?
Ans: The point estimate of µ =
5.55 minutes.
b) What is the standard error of the estimator? Interpret its
meaning.
Ans: Sx-bar = 2.22/20½ =
.49642 which indicates that the estimated value of 5.55 is closer to the true unknown
value of µ; nonetheless, a point estimate is less desirable because it fails to
account for the possibility of sampling error. Interval estimation
recognizes this error.
c) Assuming that the population has a normal distribution, develop a 95% confidence
interval for the population mean time.
Ans: .95 CI for µ = X-bar ± tcv
. Sx-bar = 5.55 ± 2.093 x .4964
Note that the degrees of freedom v = n-1 = 19 and the
margin of error (ME) = ± 2.093 x .4964 = ±1.04
Thus the P (4.51 < µ < 6.59 ) = .95
2. According to a USA Today poll (January 11, 1990), 79% of
Americans say that, if they had evidence, they would turn in a relative who
killed someone. Experts were not surprised by the poll results, saying
poll reflects " a socially desirable response" rather than real-life
action. The telephone pool of 305 adults was conducted by the Gordon S. Black
Corporation. What is the 95% confidence interval for the population proportion.
Ans:
Let X denote the # of Americans would turn in a relative who killed someone. A
random sample of size n yields P-bar = X/n = .79 (implying that
X = 305 x .79 = 241). Thus the standard error of P-bar (Óp-bar) =
[(.79x.21) /305)]½ , and the ME = ±Zcv. Óp-bar
= ± 1.96 x .0233 = ±.0457
Thus, a .95 CI for P = Pbar ± 04567 = (.7443, .8357).
3. #8.21, p. 327
The answer is: LL = 12.44 and UL = 27.96
Use the steps in #1 above
4. #8.30, p. 332
The answer is: LL = .7738 and the UL = .8262
Use the steps in #2 above
II. Hypothesis Testing Problems
1. #9.5, p. 370
Preliminary:
Let X denote weekly earnings of production workers so that µ represents
the average weekly earnings of production workers. The Bureau of Labor
Statistics, U.S. Department of Labor, wants to verify whether the average weekly
earnings of µ = 373.64 has changed since 1993. A random sample of
size n = 54 production workers is selected and their weekly earnings are
recorded. Calculated from these earning are the sample mean (X-bar =
382.13) and the sample standard deviation (S = 33.90).
Ho: µ = 373.64 which is the maintained
premise that that value has not changed since 1993.
Ha: µ is not equal to 373.64; indeed it has changed,
perhaps upward or downward since 1993.
Level of significance (alpha) = .05 which implies alpha/2 = .025 since
this is a two tailed test. Thus the Z critical value (Zcv) is Zcv
= ±1.96. The test statistic is X-bar; the CLT must be invoked to
justify its sampling distribution and thus the conversion of the value of X-bar
into Z-score, as in this case, since "n" is statistically large. Hence, for
X-bar = 382.13 the corresponding Z observed value (Zov)
= (X-bar - µ)/Sx-bar
where Sx-bar = S/n½. So Zov = (382.13 - 373.64)/4.61
= 1.84. The statistical conclusion thus is that Ho should not be rejected since
Zov is less than the Zcv. The administrative implication
of this finding is that the average earnings of the production workers have not
changed significantly from $373.64 since 1993.
2. #9.11, p. 374
3. #9.22, p. 380
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Last revised: Thursday, July 11, 2013.