Homework 7
Alizeh Azhar
10/13/2021
Problem 1: OIS Exercise 5.2, page 179 (5 pts)
For each of the following situations, state whether the parameter of interest is a mean or a proportion.
(a)
A poll shows that 64% of Americans personally worry a great deal about federal spending and the
budget deficit.
Answer: proportion
(b)
A survey reports that local TV news has shown a 17% increase in revenue within a two year period
while newspaper revenues decreased by 6.4% during this time period.
Answer: mean
(c)
In a survey, high school and college students are asked whether or not they use geolocation services on
their smart phones.
Answer: proportion
(d) In a survey, smart phone users are asked whether or not they use a web-based taxi service.
Answer: proportion
(e)
In a survey, smart phone users are asked how many times they used a web-based taxi service over the
last year.
Answer: mean
Problem 2: OIS Exercise 5.4, page 179 (7 pts)
In a random sample of 765 adults in the United States, 322 say they could not cover a $400 unexpected
expense without borrowing money or going into debt.
(a) What population is under consideration in the data set?
Answer: The population under consideration is adults in the United States.
(b) What parameter is being estimated?
Answer:
The parameter being is the population proportion of US adults who say that they could not cover a $400
unexpected expense without borrowing money or going into debt.
(c) What is the point estimate for the parameter?
Answer:
The point estimate of the population proportion of US adults who say cover a $400 unexpected expense
without borrowing money or going into debt, is the sample proportion,ˆp.
p_hat <- 322/765
p_hat
## [1] 0.420915
Therefore, ˆp =
322
765
= 0.420915
1
(d) What is the name of the statistic we can use to measure the uncertainty of the point estimate?
Answer:
The uncertainty of the point estimate can be measured by the standard error.
(e) Compute the value from part (d) for this context.
Answer:
SE
ˆp
=
q
ˆp(1−ˆp)
n
standard_error <- sqrt((p_hat*(1-p_hat))/765)
standard_error
## [1] 0.01784998
Therefore, SE
ˆp
=
q
0.421(1−0.421)
765
= 0.018
(f) A cable news pundit thinks the value is actually 50%. Should she be surprised by the data?
Answer:
If the population parameter is actually 50%, the sampling distribution would be
N
0.5,
r
0.5 × (1 − 0.5)
765
!
= N(0.5, 0.018).
pnorm(0.421, mean = 0.5, sd= sqrt(0.5*0.5/765))
## [1] 6.210504e-06
Assuming the true proportion is 50%, the probability of getting a value like 42.1% or lower is very small. The
cable news pundit should be surprised by the data.
Problem 3 (2 points)
A bottling company uses a filling machine to fill plastic bottles with cola. The bottles are supposed to contain
300 milliliters (ml). In fact, the contents vary according to a normal distribution with mean
µ
= 298 ml and
standard deviation σ = 3 ml.
(a) What is the probability that an individual bottle contains less than 295 ml?
Answer:
pnorm(295, mean = 298,sd=3)
## [1] 0.1586553
(b) What is the probability that the mean contents of the bottles in a six-pack is less than 295 ml?
Answer:
std <- 3/sqrt(6)
pnorm(295, mean = 298, sd=std)
## [1] 0.007152939
Problem 4 (2 points)
A laboratory weights filters from a coal mine to measure the amount of dust in the mine atmosphere.
Repeated measurements of the weight of dust on the same filter vary normally with standard deviation
σ
= 0
.
08 milligrams (mg) because the weighing is not perfectly precise. The dust on a particular filter actually
weighs 123 mg. Repeated weighings will then have the normal distribution with mean 123 mg and standard
deviation 0.08 mg.
2
(a) The laboratory reports the mean of 3 weighings. What is the distribution of this mean?
Answer:
The distribution of the mean is:
¯
X ∼ N
µ = 123, σ =
0.08
√
3
(b) What is the probability that the laboratory reports a weight of 124 mg or higher for this filter?
Answer:
1-pnorm(124, mean = 123, sd=0.08)
## [1] 0
Problem 5 (1 point)
The number of flaws per square yard in a type of carpet material varies with mean 1.6 flaws per square yard
and standard deviation 1.2 flaws per square yard. The population distribution cannot be normal, because a
count takes only whole-number values. An inspector studies 200 square yards of material, records the number
of flaws found in each square yard, and calculates
¯x
, the mean number of flaws per square yard inspected.
Use the Central Limit Theorem to find the approximate probability that the mean number of flaws exceeds 2
per square yard.
Answer:
1- pnorm(2,mean = 1.6, sd= 1.2/sqrt(200))
## [1] 1.214234e-06
Problem 6: OIS Exercise 5.8, page 187 (3 points)
A poll conducted in 2013 found that 52% of U.S. adult Twitter users get at least some news on Twitter.
The standard error for this estimate was 2.4%, and a normal distribution may be used to model the sample
proportion. Construct a 99% confidence interval for the fraction of U.S. adult Twitter users who get some
news on Twitter (2 points), and interpret the confidence interval in context (1 point).
Answer:
z_star <- qnorm(0.995,mean = 0, sd=1)
MoE <- z_star*0.024
0.52-MoE
## [1] 0.4581801
0.52+MoE
## [1] 0.5818199
The 99% confidence interval for the fraction of U.S. adult Twitter users who get some news on Twitter is:
(0.4581801, 0.5818199).
We are 99% confident that the true number of Twitter users in the US who get at least some news on Twitter
is between 45.8% and 58.2%.
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