Econ. 5824: Solutions, Homework # 1       Dr. Usip

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  σ =$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 (σ_x-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 σ_x-bar =  σ/n^½ provided "n" is statistically large. So  σ_x-bar = σ/n^½ [N-n/N-1]^½ . The result of the calculation is σ_x-bar = 8500/100^½ [.9664] = $821.44. 

P( X_-bar > 145,000) implies P( Z_i > Z_LL) where Z_LL = (X_-bar - µ)/σ_x-bar  = (145,000 - 137,000)/821.44 = 9.739 or 9.4. Thus P(X_-bar > 145,000) = P(Z_i > 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 (σ_p-bar ) is given as σ_p-bar = [P(1-P)/n]^½ = [(.32 x .68)/200]^½ = .0329
Thus P(P_-hat < .25) implies P(Z_i < Z_UL) where Z_UL = (P_-hat - P)/σ_p-bar  = (.25 - .32)/.0329 = -2.1276
So that P(Z_i < -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(Z_i > Z_LL) where Z_LL = (P_-hat - P)/σ_p-bar  = (.35 - .32)/.0329 = .9118
So that P(Z_i > .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(Z_i < Z_UL) where Z_UL = (P_-hat - P)/σ_p-bar  = (.70 - .68)/.0329 = .61
So that P(Z_i < .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:

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: S_x-bar = 2.22/20^½  = .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 ± t_cv . S_x-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 (σ_p-bar) = [(.79x.21) /305)]^½ , and the ME = ±Zcv. σ_p-bar  = ± 1.96 x .0233 = ±.0457 
Thus, a .95 CI for P = P_bar ± 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).

H_o: µ = 373.64 which is the maintained premise that that value has not changed since 1993.
H_a: 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 (Z_cv) is Z_cv = ±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 (Z_ov) =  (X_-bar  - µ)/S_x-bar  where S_x-bar = S/n^½. So  Z_ov = (382.13 - 373.64)/4.61 = 1.84. The statistical conclusion thus is that Ho should not be rejected since Z_ov is less than the Z_cv. 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

Top or Back to Solutions/What is New or Home page

Copyright© 1996, Ebenge Usip, all rights reserved.
Last revised: Thursday, July 11, 2013.