Rspst

CHAPTER 2

A REVIEW OF BASIC STATISTICAL CONCEPTS


ANSWERS TO ODD NUMBERED PROBLEMS

1. Descriptive Statistics

Variable N Mean Median Tr Mean StDev SE Mean
C1 28 21.32 17.00 20.69 13.37 2.53

Variable Min Max Q1 Q3

C1 5.00 54.00 11.25 28.75


a. [pic] = 21.32
b.S = 13.37
c. S2 = 178.76
d. If the policy is successful, smaller orders will be eliminated and the mean will
increase.

e. If the change causes all customers to consolidate a number of small orders into
large orders, the standard deviation will probably decrease. Otherwise, it is very
difficult to tellhow the standard deviation will be affected.

f. The best forecast over the long-term is the mean of 21.32.


3. a. Point estimate: [pic]

b. 1(( = .95 ( Z = 1.96, n = 30, [pic]
[pic]
(5.85%, 15.67%)
c. df = 30(1 = 29, t = 2.045
[pic](5.64%, 15.88%)

d. We see that the 95% confidence intervals in b and c are not much different.
This explains why a sample of size n = 30 is often taken as the cutoff between
large and small samples.



5. H0: ( = 12.1 n = 100 ( = .05
H1: ( > 12.1 S = 1.7 [pic] = 13.5

Reject H0 if Z > 1.645

Z = [pic]= 8.235

Reject H0 sincethe computed Z (8.235) is greater than the critical Z (1.645). The mean has increased.        


7. n = 60, [pic]
[pic] two-sided test, ( = .05, critical value: |Z|= 1.96
Test statistic: [pic]

Since |(2.67| = 2.67 > 1.96, reject [pic] at the 5% level. The mean satisfaction rating is
different from 5.9.
p-value: P(Z < ( 2.67 or Z > 2.67) = 2P(Z > 2.67) = 2(.0038) = .0076, very strong
evidence against [pic]


9. H0: ( = 700 n = 50 ( = .05
H1: ( ( 700 S = 50 [pic] = 715

Reject H0 if Z < -1.96 or Z > 1.96

Z = [pic] = 2.12

Since the calculated Z is greater than the critical Z (2.12 > 1.96), reject the null hypothesis. The forecast does not appear to be reasonable.p-value: P(Z < ( 2.12 or Z > 2.12) = 2 P(Z > 2.12) = 2(.017) = .034, strong evidence
against [pic]






11. a.
[pic]

b. Positive

c. (Y = 6058 (Y2 = 4,799,724 (X = 59
(X2 = 513 (XY = 48,665 r = .938




















13. This is a good population for showing how random samples are taken. Ifthree-digit random numbers are generated from Minitab as demonstrated in Problem 10, the selected items for the sample can be easily found. In this population, ( = 0.06, so most students
will not reject the null hypothesis which so states this. The few students that erroneously
reject H0 demonstrate Type I error.


15. n = 175, [pic]
Point estimate: [pic]
98% confidence interval: 1((= .98 ( Z = 2.33
[pic] ( (43.4, 47.0)

Hypothesis test:
[pic] two-sided test, ( = .02, critical value: |Z|= 2.33
Test statistic: [pic]

Since |Z| = 1.54 < 2.33, do not reject [pic] at the 2% level.

As expected, the results of the hypothesis test are consistent with the confidence
interval for (; ( = 44 is not ruled out by either procedure.CHAPTER 3

EXPLORING DATA PATTERNS AND
CHOOSING A FORECASTING TECHNIQUE


ANSWERS TO ODD NUMBERED PROBLEMS

1. Qualitative forecasting techniques rely on human judgment and intuition. Quantitative
forecasting techniques rely more on manipulation of past historical data.

3. The secular trend of a time series is the...