INFORMATION AND ANALYSIS

Answers, Ch. 8

8.9

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.

8.29

a. n = 1200, x = 912, p = x / n = 912 / 1200 = 0.76

π ≈ p ± Z

95% C.I. => Z = 1.96

=> π ≈ 0.76 ± 1.96 [0.76 * (1 - 0.76) / 1200]^0.5

=> 0.7358 < π < 0.7842

b. n = 1200, x = 672, p = x / n = 672 / 1200 = 0.56

π ≈ p ± Z

95% C.I. => Z = 1.96

=> π ≈ 0.56 ± 1.96 [0.56 * (1 - 0.56) / 1200]^0.5

=> 0.5319 < π < 0.5881

c.  Skip

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

n =

Z2 π (1 - π) ------------ e2

=

(1.96)2 (0.4372(1 - 0.372) ------------------------ (0.04)2

= 560.89 = 561

b. e = 0.04, 99% C.I. => Z = 2.576

n =

Z2 π (1 - π) ------------ e2

=

(2.576)2 (0..372)(1 - 0.372) ------------------------ (0.04)2

= 968.76 = 969

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

n =

Z2 π (1 - π) ------------ e2

=

(1.96)2 (0.372)(1 - 0.372) ------------------------ (0.02)2

= 2243.57 = 2244

d. e = 0.02, 99% C.I. => Z = 2.576

n =

Z2 π (1 - π) ------------ e2

=

(2.576)2 (0.372)(1 - 0.372) ------------------------ (0.02)2

= 3875.05 = 3876

e.  The greater the confidence level, the more observations are needed.  The smaller the acceptable sampling error, the more observations are needed.

8.63

a. e = 0.05, 95% C.I. => Z = 1.96

n =

0.25 Z2 -------- e2

=

0.25 (1.96)2 ------------ (0.05)2

= 384.15 = 385

b. n = 50, x = 42, p = x / n = 42 / 50 = 0.84

π ≈ p ± Z

95% C.I. => Z = 1.96

=> π ≈ 0.84 ± 1.96 [0.84 * (1 - 0.84) / 50]^0.5

=> 0.7384 < π < 0.9416

c.  The proportion of times the product is effective is somewhere between 74% and 94%.  The representative must decide whether that is good enough, particularly with the low end of the interval, to sell the product in the hardware stores.