This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: CHAPTER 2 A REVIEW OF BASIC STATISTICAL CONCEPTS ANSWERS TO PROBLEMS AND CASES 1. Descriptive Statistics Variable N Mean Median StDev SE Mean Orders 28 21.32 17.00 13.37 2.53 Variable Min Max Q1 Q3 Orders 5.00 54.00 11.25 28.75 a. X = 21.32 b. S = 13.37 c. S 2 = 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 tell how the standard deviation will be affected. f. The best forecast over the longterm is the mean of 21.32. 2. Descriptive Statistics Variable N Mean Median StDev SE Mean Prices 12 176654 180000 39440 11385 Variable Min Max Q1 Q3 Prices 121450 253000 138325 205625 X = 176,654 and S = 39,440 3 . a. Point estimate: % 76 . 10 = X b. 1α = .95 ⇒ Z = 1.96, n = 30, 71 . 13 , 76 . 10 = = S X ( 29 ( 29 91 . 4 76 . 10 30 / 71 . 13 96 . 1 76 . 10 / 96 . 1 ± = ± = ± n S X (5.85%, 15.67%) c. df = 30 1 = 29, t = 2.045 ( 29 ( 29 12 . 5 76 . 10 30 / 71 . 13 045 . 2 76 . 10 / 045 . 2 ± = ± = ± n S X 1 (5.64%, 15.88%) d. We see that the 95% confidence intervals in b and c are not much different because the multipliers 1.96 and 2.045 are nearly the same magnitude. This explains why a sample of size n = 30 is often taken as the cutoff between large and small samples. 4 . a. Point estimate: 63 2 59 . 102 41 . 23 = + = X 95% error margin: (102.59  23.41)/2 = 39.59 b. 1α = .90 ⇒ Z = 1.645, 2 . 20 96 . 1 / 59 . 39 / , 63 = = = n S X ( 29 23 . 33 63 ) 2 . 20 ( 645 . 1 63 / 645 . 1 ± = ± = ± n S X (29.77, 96.23) 5. H : μ = 12.1 n = 100 α = .05 H 1 : μ > 12.1 S = 1.7 X = 13.5 Reject H 0 if Z > 1.645 Z = 100 7 . 1 1 . 12 5 . 13 = 8.235 Reject H since the computed Z (8.235) is greater than the critical Z (1.645). The mean has increased. 6. point estimate: 8.1 seats interval estimate: 8.1 ± 1.96 49 7 . 5 ⇒ 6.5 to 9.7 seats Forecast 8.1 empty seats per flight; very likely the mean number of empty seats will lie between 6.5 and 9.7. 7. n = 60, 87 . , 60 . 5 = = S X 9 . 5 : 9 . 5 : 1 ≠ = μ μ H H twosided test, α = .05, critical value:  Z = 1.96 Test statistic: 67 . 2 60 / 87 . 9 . 5 60 . 5 / 9 . 5 = = = n S X Z Since  2.67 = 2.67 > 1.96, reject H at the 5% level. The mean satisfaction rating is different from 5.9. pvalue: P( Z <  2.67 or Z > 2.67) = 2 P( Z > 2.67) = 2(.0038) = .0076, very strong evidence against H . 8 . df = n 1 = 14  1 = 13, 52 . , 31 . 4 = = S X 2 4 : 4 : 1 = μ μ H H onesided test, α = .05, critical value: t = 1.771 Test statistic: 23 . 2 14 / 52 . 4 31 . 4 / 4 = = = n S X t Since 2.23 > 1.771, reject H at the 5% level. The mediumsize serving contains an average of more than 4 ounces of yogurt....
View
Full
Document
This note was uploaded on 11/07/2010 for the course BUSINESS MGT585 taught by Professor Dr.stephanies.panehaden during the Spring '08 term at Texas A&M.
 Spring '08
 DR.STEPHANIES.PANEHADEN
 Management

Click to edit the document details