In: Statistics and Probability
A random sample of
10191019
adults in a certain large country was asked "Do you pretty much think televisions are a necessity or a luxury you could do without?" Of the
10191019
adults surveyed,
522522
indicated that televisions are a luxury they could do without. Complete parts (a) through (e) below.Click here to view the standard normal distribution table (page 1).
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Click here to view the standard normal distribution table (page 2).
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(a) Obtain a point estimate for the population proportion of adults in the country who believe that televisions are a luxury they could do without.
ModifyingAbove p with caretpequals=nothing
(Round to three decimal places as needed.)
(b) Verify that the requirements for constructing a confidence interval about p are satisfied.
The sample
▼
is stated to be
can be assumed to be
cannot be assumed to be
is stated to not be
a simple random sample, the value of
▼
n ModifyingAbove p with caretnp
n ModifyingAbove p with caret left parenthesis 1 minus ModifyingAbove p with caret right parenthesisnp1−p
ModifyingAbove p with caret left parenthesis 1 minus ModifyingAbove p with caret right parenthesisp1−p
nn
ModifyingAbove p with caretp
is
nothing,
which is
▼
greater than or equal to
less than
10, and the
▼
sample size
population proportion
sample proportion
population size
▼
is stated to not be
cannot be assumed to be
can be assumed to be
is stated to be
less than or equal to 5% of the
▼
sample size.
population size.
sample proportion.
population proportion.
(Round to three decimal places as needed.)
(c) Construct and interpret a
9595%
confidence interval for the population proportion of adults in the country who believe that televisions are a luxury they could do without. Select the correct choice below and fill in any answer boxes within your choice.
(Type integers or decimals rounded to three decimal places as needed. Use ascending order.)
A.We are
nothing%
confident the proportion of adults in the country who believe that televisions are a luxury they could do without is between
nothing
and
nothing.
B.There is a
nothing%
chance the proportion of adults in the country who believe that televisions are a luxury they could do without is between
nothing
and
nothing.
(d) Is it possible that a supermajority (more than 60%) of adults in the country believe that television is a luxury they could do without? Is it likely?
It is
▼
possible, but not likely
not possible
likely
that a supermajority of adults in the country believe that television is a luxury they could do without because the
9595%
confidence interval
▼
does not contain
contains
nothing.
(Type an integer or a decimal. Do not round.)
(e) Use the results of part (c) to construct a
9595%
confidence interval for the population proportion of adults in the country who believe that televisions are a necessity.The
9595%
confidence interval is
(nothing,nothing).
(Round to three decimal places as needed.)
Click to select your answer(s).
Please answer all
a)
point estimate for the population proportion =p̂ = x/n = 522/1019 = 0.512
b)
The sample (is stated to be ) a simple random sample,
the value of (np^(1-p^)
is 254.597 which is (greater than or equal
to)10 and the sample size can be (assumed to be) less than or equal
to 5% of (population size).
c)
Level of Significance, α =
0.05
Number of Items of Interest, x =
522
Sample Size, n = 1019
Sample Proportion , p̂ = x/n =
0.512
z -value = Zα/2 = 1.960 [excel
formula =NORMSINV(α/2)]
Standard Error , SE = √[p̂(1-p̂)/n] =
0.0157
margin of error , E = Z*SE = 1.960
* 0.0157 = 0.0307
95% Confidence Interval is
Interval Lower Limit = p̂ - E = 0.512
- 0.0307 = 0.4816
Interval Upper Limit = p̂ + E = 0.512
+ 0.0307 = 0.5430
95% confidence interval is (
0.482 < p < 0.543
)
We are 95%
confident the proportion of adults in the country who believe that televisions are a luxury they could do without is between 0.482 and 0.543
d)
it is (possible but not likely )
that a supermajority of adults in the country believe that television is a luxury they could do without because the
95%
confidence interval (does not contain 0.60)
e)
Level of Significance, α =
0.05
Number of Items of Interest, x =
497
Sample Size, n = 1019
Sample Proportion , p̂ = x/n =
0.488
z -value = Zα/2 = 1.960 [excel
formula =NORMSINV(α/2)]
Standard Error , SE = √[p̂(1-p̂)/n] =
0.0157
margin of error , E = Z*SE = 1.960
* 0.0157 = 0.0307
95% Confidence Interval is
Interval Lower Limit = p̂ - E = 0.488
- 0.0307 = 0.4570
Interval Upper Limit = p̂ + E = 0.488
+ 0.0307 = 0.5184
95% confidence interval is (
0.457 < p < 0.518
)