STA 3032, Fall 2014
Solutions to Homework #11
HW #11: pp. 214-215: 4.74, 4.78a, 4.80abc
We will discuss these in class, if needed.
4.74. Of 1000 randomly selected cases of lung cancer, 823 resulted in death.
a) Test the hypothesis 𝐻0 : 𝑝 = 0.85 v. 𝐻𝐴 : 𝑝 ≠ 0.85, with α = 0.05.
Step 1: 𝐻0 : 𝑝 = 0.85 𝐻𝐴 : 𝑝 ≠ 0.85
Step 2: n = 1000, α = 0.05.
𝑝̂−0.85
Step 3: The test statistic is 𝑍 = (0.85)(0.15), which under H0 has an approximate standard normal
√
1000
distribution.
Step 4: We will reject H0 if the p-value is less than 0.05.
Step 5: We have z = -2.3912, and p-value = 0.016795.
Step 6: We reject H0 at the 0.05 level of significance. We have sufficient evidence to conclude that
the proportion of lung cancer patients who die is not 85%.
b) Find a 95% CI for the death rate from lung cancer.
We are 95% confident that the death rate from lung cancer is between 79.934% and 84.666%.
OR
We are 95% confident that the death rate from lung cancer is 82.3% ± 2.366%.
c) How large a sample would be required to be at least 95% confident that the error in estimating the
death rate from lung cancer is less than 3%?
𝑧𝛼 2 1
1.9599 2 1
2
𝑛 = ( ) ↑= (
) ↑= 1068.
𝐸 4
0.03
4
4.78. A manufacturer of electronic calculators is interested in estimating the fraction of defective units
produced. A random sample of 800 calculators contains 10 defectives.
a) Step 1: 𝐻0 : 𝑝 ≤ 0.01
Step 2: n = 800, α = 0.05.
𝐻𝐴 : 𝑝 > 0.01.
Step 3: The test statistic is 𝑍 =
𝑝̂−0.01
, which under H0 has an approximate standard normal
(0.01)(0.99)
√
800
distribution.
Step 4: We will reject H0 if z > 1.645 𝑜𝑟 if the p-value is less than 0.05.
Step 5: We have z = 0.710669, and p-value = 0.238645.
Step 6: We fail to reject H0 at the 0.05 level of significance. We do not have sufficient evidence to
conclude that the proportion of defective units exceeds 1%.
4.80. The fraction of defective integrated circuits produced in a photolithography process is being
studied. A random sample of 300 circuits is tested, revealing 18 defectives.
a) Step 1: 𝐻0 : 𝑝 = 0.04
𝐻𝐴 : 𝑝 ≠ 0.04.
Step 2: n = 300, α = 0.05.
𝑝̂−0.04
Step 3: The test statistic is 𝑍 = (0.04)(0.96), which under H0 has an approximate standard normal
√
300
distribution.
Step 4: We will reject H0 if |𝑧| > 1.96 𝑜𝑟 if the p-value is less than 0.05.
Step 5: We have z = 1.7677, and p-value = 0.0771.
Step 6: We fail to reject H0 at the 0.05 level of significance. We do not have sufficient evidence to
conclude that the proportion of defective circuits is not 1%.
b) p-value = 0.0771.
c) We are 95% confident that the proportion of defective circuits in the population is between 3.313%
and 8.687%.
OR
We are 95% confident that the proportion of defective circuits in the population is 6.0% ± 2.687%.