NOTE: Use t-distribution, when n is less than 30, or ? unknown or both cases happen.

1. When a 99% confidence interval is calculated instead of a 95% confidence interval with n being the same, the maximum error of estimate will be

(a) Smaller

(b) larger

(c) the same

(d) it cannot be determined

2. A random sample of 49 shoppers showed that they spend an average of $23.45 per visit at a mall bookstore. The standard deviation of the population is $2.80. Find the 90% confidence interval of the true mean.

Solution:n= 49, x=23.45, s=2.8, a=1-0.9=0.1

90% CI for the mean (s known) = x±z(α2)σn

z(a/2)=z(0.05)=1.645

Answer: 23.45±1.6452.849=(22.792,24.108)

3. The average weigh of 40 randomly minivans was 4150 pounds. The standard deviation was 480 pounds. Find the 99% confidence interval of the true mean weight of the minivans.

Solution:n= 40, x=4150, s=480, a=1-0.99=0.01

99% CI for the mean (s unknown) = x±t(α2,n-1)sn

t(a/2,n-1)=t(0.005,39)=2.708

Answer: 4150±2.70848040=(3944.477, 4355.523)

4. In a study of 10 insurance sales representatives from a certain large city, the average age of the group was 48.6 years and the standard deviation was 4.1 years. Find the 95% confidence interval of the population mean age of all insurance sales representatives in that city.

Solution:n= 10, x=48.6, s=4.1, a=1-0.95=0.05

95% CI for the mean (s unknown) = x±t(α2,n-1)sn

t(a/2,n-1)=t(0.025,9)=2.262

Answer: 48.6±2.2624.19=(45.667, 51.533)

5. In a hospital, a sample of 8 weeks was selected, and it was found that an average of 438 patients was treated in the emergency room each week. The standard deviation was 16. Find the 99% confidence interval of the true mean.

Solution:n= 8, x=438, s=16, a=1-0.99=0.01

99% CI for the mean (s unknown) = x±t(α2,n-1)sn

t(a/2,n-1)=t(0.005,7)=3.499

Answer: 438±3.499168=(418.207, 457.793)

6. A university dean wishes to estimate the average number of hours that the freshmen study each week. The standard deviation from a previous study is 2.6 hours. How large a sample must be selected if he wants to be 99% confident of finding whether the true mean differs from a sample mean by 0.5 hour?

Solution: s=2.6, a=1-0.99=0.01

Maximum error of estímate = zα2σn=z0.0052.6n=0.5

z(0.005)=2.576

2.5762.6n=0.5→n=2.5762.60.5→n=2.5762.60.52=179.4

Answer: we need a simple size of 180

7. A recent study of 75 workers found that 53 people rode the bus to work each day. Find the 95% confidence interval of the proportion of all workers who rode the bus to work.

Solution: n= 75, x=53, p=xn=53/75, a=1-0.95=0.05

Ci at 95% for proportion is: p±z(α2)p(1-p)n

z(a/2)=z(0.025)=1.96

Answer: 53/75±1.9653/75(1-53/75)75=(0.604, 0.81)

8. In a study of 150 accidents that required treatment in an emergency room, 36% involved children under 6 years of age. Find the 90% confidence interval of the true proportion of accidents that involve children under the age of 6.

Solution: n= 150, p=0.36, a=1-0.9=0.1

Ci at 90% for proportion is: p±z(α2)p(1-p)n

z(a/2)=z(0.05)=1.645

Answer: 0.36±1.6450.36(1-0.36)150=(0.296, 0.424)

9. In a survey of 1004 individuals, 442 felt that President Barack Obama spent too much time away from Washington. Find a 95% confidence interval for the true proportion.

Solution: n= 1004, x=442, p=xn=442/1004, a=1-0.95=0.05

Ci at 95% for proportion is: p±z(α2)p(1-p)n

z(a/2)=z(0.025)=1.96

Answer: 442/1004±1.96442/1004(1-442/1004)1004=(0.41, 0.471)

10. The data below represent children’s animated film lengths in minutes. Estimate the true length of all children’s animated films with 95% confidence. 93 83 76 92 77 81 78 100 78 76 75

Solution:n= 11, x=82.636, s=8.489, a=1-0.95=0.05

95% CI for the mean (s unknown) = x±t(α2,n-1)sn

t(a/2,n-1)=t(0.025,10)=2.228

Answer: 82.636±2.2288.48911=(77.933, 88.339)