HW10 Solutions
7.4
- 25.9
1.96*2.7/sqrt(90) =
25.9
0.56 =
(25.34, 26.46)
- 25.9
1.645*2.7/sqrt(90) =
25.9
0.47=
(25.43, 26.37)
- 25.9
2.576*2.7/sqrt(90) =
25.9
0.73=
(25.17, 26.63)
7.6
- 28
1.96*sqrt(12)/sqrt(75) =
28
0.78
(27.22, 28.78)
- 102
1.96*sqrt(22)/sqrt(200) =
102
0.65
(101.35, 102.65)
- 15
1.96*0.3/sqrt(100) =
15
0.06
(14.94, 15.06)
- 4.05
1.96*0.83/sqrt(100) =
4.05
0.16
(3.89, 4.21)
- No. For large sample sizes it does not matter what distribution the
parent population has, the Central Limit Theorem tells us that the
sampling distribution will be normal regardless.
7.21
- If the parent population is normally distributed then the sampling
distribution of
will be the normal distribution regardless of sample size.
- If we don't know the distribution of the parent population then we
don't know what the sampling distribution of
would be when the sample size is small. However, if the sample size is
large then the Central Limit Theorem tells us that the sampling
distribution of
will be the normal distribution regardless of the parent population's
distribution.
7.22
- t = 1.440; z = 1.282
- t = 1.943; z = 1.645
- t = 2.447; z = 1.96
- t = 3.143; z = 2.326
- t = 3.707; z = 2.576
7.26 (the sample mean is 98 and the sample variance is 160 and n=16)
- 98
1.341*sqrt(160)/sqrt(16) =
98
4.24
(93.76, 102.24)
- 98
2.131*sqrt(160)/sqrt(16) =
98
6.74
(91.26, 104.74)
The 95% confidence interval is wider.
- I am 80% confident that the population mean is between 93.76 and
102.24 and I'm 95% confident that the population means is between
91.26 and 104.74. By "confident" I mean that if I were to
take repeated random samples from the population and calculate
confidence intervals from each sample then 80% of 80% CIs and 95% of
95% CIs would actually contain the population mean.
The 95% CI is wider than the 80% CI because I'm more confident that
the population mean would be in a larger set of numbers than in a
smaller set. The smaller the set of numbers (or the width of the
interval) the higher the chance that the population mean is not in the
interval and so the lower the confidence level.
7.35
When n is large the central limit theorem tells us that 
is normally distributed because
is a special form of .
The sampling distribution of
is p and the variance of the sampling distribution is p(1 - p)/n.
7.36
is an
unbiased estimator of p because the mean of the sampling
distribution of p ( )
equals p.
7.38
- I shouldn't have assigned this problem and I won't grade it. Sorry.
- 0.76
1.645*sqrt[0.76*(1 - 0.76) /144]
0.76
0.059
(0.701, 0.819)
- We only need to assume that the sample size is large enough for the
Central Limit Theorem to hold and that the sample was randomly
selelcted.
Also answer the following questions:
- Suppose you took a random sample of 500 people and you asked if
they have filled out their census form yet.
is the proportion of the sample that said yes, they filled out the
form. If 80% of the population have actually filled out the census
(meaning p = 0.8) the what is the probability
P(0.75 < < 0.85)?
(Remember that the Central Limit Theorem tells us that
is normally distributed with mean p and variance p(1-p)/n
for large n.)
Answer:
P(0.75 < < 0.85) =
P{(0.75 - 0.8)/sqrt[0.8*(1-0.8)/500] < Z < (0.85 - 0.8)/sqrt[0.8*(1-0.8)/500]} =
P(-2.80 < Z < 2.80) =
1 - 2*0.0026 =
0.9948
- What is the difference between a sample standard deviation and a
sample standard error? What do they estimate?
Answer:
The sample standard deviation is used to estimate the variability of
the parent population. The sample standard error is used to estimate
the variability of the sampling distribution.
|
|
