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Study Set
Essentials of Statistics Study Set 1
Quiz 7: Estimates and Sample Size
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Question 21
Multiple Choice
50 people are selected randomly from a certain population and it is found that 18 people in the sample are over 6 feet tall. What is the point estimate of the proportion of people in the population who are over 6 feet tall?
Question 22
Multiple Choice
A
99
%
99 \%
99%
confidence interval (in inches) for the mean height of a population is
66.2
<
μ
<
67.6
66.2 < \mu < 67.6
66.2
<
μ
<
67.6
. This result is based on a sample of size 144 . Construct the
95
%
95 \%
95%
confidence interval. (Hint: you will first need to find the sample mean and sample standard deviation) .
Question 23
Multiple Choice
Use the given degree of confidence and sample data to construct a confidence interval for the population mean µ. Assume that the population has a normal distribution. -A sociologist develops a test to measure attitudes towards public transportation, and 27 randomly selected subjects are given the test. Their mean score is 76.2 and their standard deviation is 21.4. Construct the 95% confidence interval for the mean score of all such subjects.
Question 24
Multiple Choice
Use the confidence level and sample data to find a confidence interval for estimating the population µ. Round your answer to the same number of decimal places as the sample mean. -Test scores:
n
=
79
,
x
‾
=
65.0
,
σ
=
5.0
;
98
%
\mathrm { n } = 79 , \overline { \mathrm { x } } = 65.0 , \sigma = 5.0 ; 98 \%
n
=
79
,
x
=
65.0
,
σ
=
5.0
;
98%
confidence
Question 25
Essay
The Bide-a-While efficiency hotel, which caters to business workers who stay for extended periods of time (weeks or months), offers room service. In a small study of 35 randomly selected room service orders, the 95% confidence interval for mean delivery time for room service is
24.8
<
μ
<
29.6
24.8 < \mu < 29.6
24.8
<
μ
<
29.6
minutes. The marketing director is trying to determine if she can advertise "room service in under 30 minutes, or the order is free." How would you advise her?
Question 26
Multiple Choice
Use the confidence level and sample data to find a confidence interval for estimating the population µ. Round your answer to the same number of decimal places as the sample mean. -A group of 64 randomly selected students have a mean score of
38.6
38.6
38.6
with a standard deviation of
4.9
4.9
4.9
on a placement test. What is the
90
%
90 \%
90%
confidence interval for the mean score,
μ
\mu
μ
, of all students taking the test?
Question 27
Essay
Under what three conditions is it appropriate to use the t distribution in place of the standard normal distribution?
Question 28
Multiple Choice
Use the given data to find the minimum sample size required to estimate the population proportion. -Margin of error: 0.018; confidence level:
99
%
;
p
^
and
q
^
unknown
99 \% ; \hat { \mathrm { p } } \text { and } \hat { \mathrm { q } } \text { unknown }
99%
;
p
^
and
q
^
unknown
Question 29
Multiple Choice
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. -
n
=
110
,
x
=
68
;
88
%
\mathrm { n } = 110 , \mathrm { x } = 68 ; 88 \%
n
=
110
,
x
=
68
;
88%
confidence
Question 30
Multiple Choice
In constructing a confidence interval for
σ
\sigma
σ
or
σ
2
\sigma ^ { 2 }
σ
2
, a table is used to find the critical values
χ
L
2
\chi _ { L } ^ { 2 }
χ
L
2
and
χ
2
R
\chi \underset { R } { 2 }
χ
R
2
for values of
n
≤
101
n \leq 101
n
≤
101
. For larger values of
n
,
χ
L
2
n , \chi _ { L } ^ { 2 }
n
,
χ
L
2
and
χ
R
2
\chi { } _ { \mathrm { R } } ^ { 2 }
χ
R
2
can be approximated by using the following formula:
χ
2
=
1
2
[
±
z
α
/
2
+
2
k
−
1
]
2
\chi ^ { 2 } = \frac { 1 } { 2 } [ \pm \mathrm { z } \alpha / 2 + \sqrt { 2 \mathrm { k } - 1 } ] ^ { 2 }
χ
2
=
2
1
[
±
z
α
/2
+
2
k
−
1
]
2
where
k
\mathrm { k }
k
is the number of degrees of freedom and
z
α
/
2
\mathrm { z } \alpha / 2
z
α
/2
is the critical z score. Construct the
90
%
90 \%
90%
confidence interval for
σ
\sigma
σ
using the following sample data: a sample of size
n
=
234
\mathrm { n } = 234
n
=
234
yields a mean weight of
155
l
b
155 \mathrm { lb }
155
lb
and a standard deviation of
26.0
l
b
26.0 \mathrm { lb }
26.0
lb
. Round the confidence interval limits to the nearest hundredth.
Question 31
Essay
A radio show host asked people to call in and say whether they support new legislation to promote cleaner sources of energy. Based on this sample, she constructed a confidence interval to estimate the proportion of all listeners to her show who support the legislation. Is the confidence interval likely to give a good estimate of the proportion of her listeners who support the legislation?
Question 32
Essay
A paper published the results of a poll. It stated that, based on a sample of 1000 married men, 51% of married men say that they would marry the same woman again. The margin of error was given as ±3 percentage points and the confidence level was given as 95%. What does it mean that the margin of error was ±3 percentage points?
Question 33
Multiple Choice
Use the given data to find the minimum sample size required to estimate the population proportion. -Margin of error: 0.008; confidence level: 99%; from a prior study,
p
^
\hat{p}
p
^
is estimated by 0.208.
Question 34
Essay
Define a point estimate. What is the best point estimate for
μ
\mu
μ
?
Question 35
Essay
Mark wanted to estimate the mean number of years of education of adults in his city. He waited outside a public library and interviewed every tenth adult leaving. Based on this sample, he constructed a confidence interval for the mean number of years of education of adults in the city. Do you think this confidence interval will give a good estimate? Why or why not?
Question 36
Essay
Draw a diagram of the chi-square distribution. Discuss its shape and values.
Question 37
Essay
Define confidence interval and degree of confidence. Make up an example of a confidence interval and interpret the result.
Question 38
Multiple Choice
Of 366 randomly selected medical students, 27 said that they planned to work in a rural community. Find a 95% confidence interval for the true proportion of all medical students who plan to work in a rural community.