**I. Sampling Distribution Problems
1. #7.19, p. 292
**Preliminary:

**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 (Ó_{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 Ó_{x-bar}
= Ó/n^{½} provided "n" is statistically large.
So Ó_{x-bar} = Ó/n^{½ } [N-n/N-1]^{½} . The
result of the calculation is Ó_{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)/**Ó_{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 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 (Ó_{p-bar}
) is given as Ó_{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**)/**Ó_{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**)/**Ó_{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**)/**Ó_{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:

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: S

c) Assuming that the population has a normal distribution, develop a 95% confidence interval for the population mean time.

Thus the P (4.51 <

2

Ans:

Let X denote the # of Americans would turn in a relative who killed someone. A random sample of size n yields P

Thus, a .95 CI for P = P

**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

**H _{o}**:

Level of significance (alpha) = .05 which implies alpha/2 = .025 since this is a two tailed test. Thus the Z critical value (Z

**2. #9.11, p. 374**

**3. #9.22, p. 380**

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