1. No category

Solution:
a) Population: people who recently bought new cars
Sample: the 167 people who were surveyed.
̂
this is the proportion that comes from the sample
this is the actual proportion of people who expressed dissatisfaction for the whole population
b) Population: college students
Sample: the 2883 students who were surveyed.
̂
this is the proportion that comes from the sample
this is the actual proportion of college students who have cell phones
c) Population: potato plants in a field in Maine
Sample: the 240 plants that were checked
̂
this is the proportion that comes from the sample
this is the actual proportion of plants that were affected by blight for the whole field
d) Population: any employees of this company (current and future)
Sample: the 309 current employees
̂
this is the proportion of employees that suffered an injury last year
this is the proportion of injuries for ALL employees of this company.
Solution:
The formula for finding a confidence interval is:
̂
̂
a) Correct.
̂̂
√ , where za depends on the confidence level.
√
̂̂
. If you keep n the same, me gets smaller only if za gets smaller. Smaller za
means a narrower confidence.
̂̂
b) Correct.
√ . If you keep za the same, then increasing n decreases the margin of error.
c) Correct.
√ . If you decrease n, you also have to decrease za in order to keep me the
̂̂
same.
d) Correct.
√
̂̂
√
̂̂
How to find the za value for a specified confidence interval:
In the formula for margin of error, or the confidence interval we need to know the value of za. This value
is dependent on the confidence level.
Here is an example of how it is found for 90% confidence.
95% confidence means that there is a probability of .9000 under the normal distribution curve between
the values of –za and za:
Subtract this from 1:
1 - .9000 = .1 and divide by 2,
We find that there is a probability of .05 in each of the tails of the distribution.
To find the value of za notice that we have a probability of .9000 + .05 = .9500 to the left of the za value.
Find the closest value to .9500 and read the z-value associated to that.
The z-value is halfway between 1.64 and 1.65, so za = 1.645 which is often rounded to za = 1.65
Solution:
CI = ̂
√
̂̂
Here, n = 525, ̂
This means CI =
̂
, za = 1.96
√
(
)
)
(
)
b) We are 95% confident that between 80% and 86% of all the broiler chickens sold in the US were
infected with Campylobacter.
c) The size of the population is irrelevant. If you choose random samples 95% of intervals generated by
these samples will capture the true contamination level.
Solution:
CI = ̂
√
̂̂
Here, n = 1003, ̂
̂
, for 95% confidence: za = 1.96, for 90% confidence: za = 1.65
me = 0.018
√
This means CI =
(
)(
)
(
)
b) We are 95% confident that between 9.2% and 12.8% of adults approve of attempts of cloning
humans.
c) The margin of error will be smaller. me = 0.016
Here CI =
(
√
)(
)
(
)
e) Smaller samples will produce larger margin of errors.
Solution:
a) CI = ̂
√
̂̂
Here, n = 582, ̂
̂
, for 95% confidence: za = 1.96
me = 0.03
This means CI =
√
(
)(
)
(
)
b) We are 95% confident, based on this sample, that between 13% and 19% of all accidents involve
teenage drivers.
c) About 95% of all random samples will produce confidence intervals that contain the true population
proportion.
d) Contradicts. The interval is completely below 20% (1 in 5).
Solution:
b) CI = ̂
√
Here, n = 2700, ̂
̂̂
̂
, for 98% confidence: za = 2.33
me = 0.018
This means CI =
√
(
)(
)
(
)
b) We are 98% confident, based on this sample, that between 18.2% and 21.8% of English children are
deficient in vitamin D.
c) About 98% of all random samples will produce confidence intervals that contain the true population
proportion.
Solution:
√
a) CI = ̂
̂̂
Here, n = 153, ̂
̂
, for 90% confidence: za = 1.65
me = 0.054
√
This means CI =
(
)(
)
(
)
b) me = 0.054 but we want it to be .027
√
(
̂̂
(
√
)(
)
(
√
)(
)
)
(
)(
√
)
(
(
(
)(
)(
)
)
)
c) Sample might not be random or representative. Deer that may be legally hunted may not be
representative of all sexes and ages.
√
Here, n = 972, ̂
√
(
)(
, for ?? confidence: za = ??
̂̂
√
√
̂
̂̂
(
)
)(
)
63
To find the confidence level find the probability in the normal distribution curve between
z = -1.63 and z = 1.63
Find the probability to the left of z = -1.63. From the table we get .0516.
This means that there is a probability of .0516 to the left of z = -1.63 and there is a probability to
the right of z = 1.63.
The probability in the middle region is: 1 – 2(.0516) = 0.8968.
So, this represents a confidence interval of 89.7% OR 90%