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, 𝑥̅ =23.45, 
𝜎
90% CI for the mean ( known) = 𝑥̅ ± 𝑧(𝛼⁄2) 𝑛
√
z(/2)=z(0.05)=1.645
Answer: 23.45 ± 1.645
2.8
√49
= (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, 𝑥̅ =4150, s
99% CI for the mean ( unknown) = 𝑥̅ ± 𝑡(𝛼⁄2 , 𝑛 − 1)
𝑠
√𝑛
t(/2,n-1)=t(0.005,39)=2.708
Answer: 4150 ± 2.708
480
√40
= (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, 𝑥̅ =48.6, s
95% CI for the mean ( unknown) = 𝑥̅ ± 𝑡(𝛼⁄2 , 𝑛 − 1)
𝑠
√𝑛
t(/2,n-1)=t(0.025,9)=2.262
Answer: 48.6 ± 2.262
4.1
√9
= (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, 𝑥̅ =438, s
99% CI for the mean ( unknown) = 𝑥̅ ± 𝑡(𝛼⁄2 , 𝑛 − 1)
𝑠
√𝑛
t(/2,n-1)=t(0.005,7)=3.499
Answer: 438 ± 3.499
16
√8
= (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: 
𝜎
2.6
Maximum error of estímate = 𝑧(𝛼⁄2) 𝑛 = 𝑧(0.005) 𝑛 = 0.5
√
√
z(0.005)=2.576
2.576
2.6
√𝑛
= 0.5 → √𝑛 = 2.576
2.6
2.6 2
→ 𝑛 = (2.576 ) = 179.4
0.5
0.5
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.
𝑥
Solution: n= 75, x=53, 𝑝̂ = 𝑛 = 53/75, =1-0.95=0.05
𝑝̂(1−𝑝̂)
𝑛
Ci at 95% for proportion is: 𝑝̅ ± 𝑧(𝛼⁄2)√
z(a/2)=z(0.025)=1.96
53/75(1−53/75)
75
Answer: 53/75 ± 1.96√
= (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, 𝑝̂ = 0.36, =1-0.9=0.1
𝑝̂(1−𝑝̂)
𝑛
Ci at 90% for proportion is: 𝑝̅ ± 𝑧(𝛼⁄2)√
z(a/2)=z(0.05)=1.645
0.36(1−0.36)
150
Answer: 0.36 ± 1.645√
= (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, 𝑝̂ =
𝑥
𝑛
= 442/1004, =1-0.95=0.05
𝑝̂(1−𝑝̂)
𝑛
Ci at 95% for proportion is: 𝑝̅ ± 𝑧(𝛼⁄2)√
z(a/2)=z(0.025)=1.96
442/1004(1−442/1004)
1004
Answer: 442/1004 ± 1.96√
= (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, 𝑥̅ =82.636, s
95% CI for the mean ( unknown) = 𝑥̅ ± 𝑡(𝛼⁄2 , 𝑛 − 1)
t(/2,n-1)=t(0.025,10)=2.228
Answer: 82.636 ± 2.228
8.489
√11
= (77.933, 88.339)
𝑠
√𝑛