STAT500 HW#6_solution
Stat 500 Solution- HW6
1) (0 pts)
a) The population of interest is the lifetime of all of the fuses of that particular kind produced
by that manufacturer.
b) The answer to this question is involved in a hypothesis testing because we want to use
sample average to know whether the population average is as much as the manufacturer
claimed.
2) (25 pts)
It is okay to use the 1-proportion z-interval because: (83/120)*120 = 0.692*120 = 83 > 5
AND (1-83/120)*120 = 0.308*120 = 37 > 5.
a. 95% CI
1 – α = 0.95  α = 0.05, α/2 = 0.025  zα/2 = z0.025 = 1.96
95% confidence interval:
ˆ  z0.025 ˆ (1  ˆ ) / n . (Square root is over the entire right hand
side)
0.692  1.96*(0.0421) = 0.692± 0.0825= (0.609484, 0.774516)
Interpretation of the interval: we are 95% confident that the true proportion of individuals
knowledgeable of the product in the population is between 0.609 and 0.774.
b. 99% CI
1 – α = 0.99  α = 0.01, α/2 = 0.005  zα/2 = z0.005 = 2.575
𝜶
̂ ± 𝒁 ∗ √𝝅
̂ (𝟏 − 𝝅
̂ )/𝒏
99% confidence interval: 𝝅
𝟐
(square taken over the entire right hand
side)
0.692  2.575*(0.0421) = 0.692  0.1084075= (0.5835925, 0.8004075)
Interpretation of the interval: we are 99% confident that the true proportion of individuals
knowledgeable of the product in the population is between 0.584 and 0.800.
The margin of error for the 95% confidence interval is 0.0825 and the margin of error for the
99% confidence interval is 0.1084075. The 99% confidence interval has a larger margin of
error than the 95% confidence interval, which means that the level of precision is lower.
3)
(25 pts)
95% Confidence Interval Exact Interval:
Exact
Sample
1
X
83
N
120
Sample p
95% CI
P-Value
0.691667 (0.600863, 0.772745)
0.000
95% Confidence Interval 1-proportion z-interval:
Sample
1
X
83
N
120
Sample p
95% CI
Z-Value P-Value
0.691667 (0.609041, 0.774293)
1
-12.98
0.000
STAT500 HW#6_solution
99% Confidence Interval Exact Interval:
Exact
Sample
1
X
83
N
120
Sample p
99% CI
P-Value
0.691667 (0.572800, 0.794870)
0.000
99% Confidence Interval 1-proportion z-interval:
Sample
1
X
83
N
120
Sample p
99% CI
Z-Value P-Value
0.691667 (0.583078, 0.800255)
-32.85
0.000
4) (20pts)
Use the Conservative Method to guarantee that the sample size will be large enough. ˆ = 0.5,
E = 0.04; (1-α) = 0.93, α/2 = 0.035, zα/2 = z0.035 = 1.812
 n = (zα/2)2 (0.5)(0.5) /E2 = [(1.812)2(0.5)(0.5)]/(0.04) 2 = 512.9719
So, 513 people are needed to guarantee that the sample size will be large enough.
Using sample size by conservative method has no risk that CI may be too wide. So we would
use conservative answer for this problem (plus we want to be guaranteed that it is large enough
and not have to go back and sample again).
5) (20pts)
In this problem since the set up cost of sampling additional people later on is relatively small
compared to sampling extra units, we would use the educated guess.
Use the Educated Guess Method,. ˆ = 0.692, E = 0.04; (1-α) = 0.93, α/2 = 0.035, zα/2 = z0.035 =
1.812
 n = (zα/2)2 ˆ (1- ˆ )/E2 = [(1.812)2(0.692)(1-0.692)]/(0.04)2 = 437.37
Then 438 people are needed.
6) (0 pts)
Because 9*(4/9) = 4 and 9*( 1- (4/9) )= 5 we cannot use the normal approximation. Instead, we
use the exact proportions to derive the confidence interval. Using Minitab, we have the
following result.
Exact
90% CI for One Proportion
Exact
Sample X N Sample p
1
4
9
0.444444
90% CI
(0.168750, 0.748632)
P-Value
0.001
Hence, we can say that there is 90% confidence that the proportion of people wearing eyeglasses
will fall between 0. 168750 and 0.748632.
2