STATISTICS FOR MANAGERS

Answers, Ch. 8

8.7

σ = 0.02, n = 50, = 0.995

Note: The square root of n will be designated as n^0.5.

a. 99% C.I. => Z = 2.576

=> m 0.995 ± 2.576 (0.02 / 50^0.5)

=> 0.988 < m < 1.002

b. No, the claimed amount of paint falls within the confidence interval

c. No, since n = 50 > 30, the Central Limit Theorem can be invoked

d. 95% C.I. => Z = 1.96

=> m 0.995 ± 1.96 (0.02 / 50^0.5)

=> 0.989 < m < 1.001

8.17

n = 18, = 195.3, s = 21.4

a.

95% C.I. => t_18-1 = 2.1098

=> m 195.3 ± 2.1098 (21.4 / 18^0.5)

=> 184.66 < m < 205.94

b. No, the claimed performance is in the calculated confidence interval

c. It is not unusual for one tire to be at 210, as that would be within one standard deviation of the mean.  It would be unusual for the average of 18 tires to be that high.

8.31

a. n = 500, x = 370, p = x / n = 370 / 500 = 0.74

π ≈ p ± Z

95% C.I. => Z = 1.96

=> π ≈ 0.74 ± 1.96 [0.74 * (1 - 0.74) / 500]^0.5

=> 0.702 < π < 0.778

b. n = 500, x = 370, p = x / n = 370 / 500 = 0.74

π ≈ p ± Z

99% C.I. => Z = 2.576

=> π ≈ 0.74 ± 2.576 [0.74 * (1 - 0.74) / 500]^0.5

=> 0.689 < π < 0.791

c.  The 99% C.I. is wider.  The interval has to be wider to have a greater probability of including the population mean.

8.43

σ = 45, e = 5

a. 90% C.I. => Z = 1.645

n =

Z2 σ2 ------ e2

=

(1.645)2 (45)2 --------------- (5)2

= 219.188 = 220

b. 99% C.I. => Z = 2.576

n =

Z2 σ2 ------ e2

=

(2.576)2 (45)2 --------------- (5)2

= 537.50 = 538

8.45

a. e = 0.06, 95% C.I. => Z = 1.96

n =

Z2 π (1 - π) ------------ e2

=

(1.96)2 (0.1577)(1 - 0.1577) ------------------------ (0.06)2

= 141.75 = 142

b. e = 0.04, 95% C.I. => Z = 1.96

n =

Z2 π (1 - π) ------------ e2

=

(1.96)2 (0.1577)(1 - 0.1577) ------------------------ (0.04)2

= 318.93 = 319

c. e = 0.02, 95% C.I. => Z = 1.96

n =

Z2 π (1 - π) ------------ e2

=

(1.96)2 (0.1577)(1 - 0.1577) ------------------------ (0.02)2

= 1275.71 = 1276