Question

In a previous​ year,

5656​%

of females aged 15 and older lived alone. A sociologist tests whether this percentage is different today by conducting a random sample of

750750

females aged 15 and older and finds that

410410

are living alone. Is there sufficient evidence at the

alphaαequals=0.10.1

level of significance to conclude the proportion has​ changed?

Because np 0 left parenthesis 1 minus p 0 right parenthesisnp01−p0equals=nothing

▼

​10, the sample size is

▼

​5% of the population​ size, and the sample

▼

the requirements for testing the hypothesis

▼

are not

are

satisfied.

​(Round to one decimal place as​ needed.)

Identify the null and alternative hypotheses for this test.

Upper H 0H0​:

pp

equals=

. 56.56 versus

Upper H 1H1​:

pp

not equals≠

. 56.56

​(Type integers or decimals. Do not​ round.)

Find the test statistic for this hypothesis test.

z 0z0equals=nothing

​(Round to two decimal places as​ needed.)

Determine the​ P-value for this hypothesis test.

​P-valueequals=nothing

​(Round to three decimal places as​ needed.)

State the conclusion for this hypothesis test.

A.

RejectReject

Upper H 0H0.

There

is notis not

sufficient evidence at the

alphaαequals=0.10.1

level of significance to conclude that the proportion of females who are living alone has changed.

B.

RejectReject

Upper H 0H0.

There

isis

sufficient evidence at the

alphaαequals=0.10.1

level of significance to conclude that the proportion of females who are living alone has changed.

C.

Do not rejectDo not reject

Upper H 0H0.

There

isis

sufficient evidence at the

alphaαequals=0.10.1

level of significance to conclude that the proportion of females who are living alone has changed.

D.

Do not rejectDo not reject

Upper H 0H0.

There

is notis not

sufficient evidence at the

alphaαequals=0.10.1

level of significance to conclude that the proportion of females who are living alone has changed.

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p = 0.56
Alternative Hypothesis, Ha: p ≠ 0.56

Test statistic,
z = (pcap - p)/sqrt(p*(1-p)/n)
z = (0.5467 - 0.56)/sqrt(0.56*(1-0.56)/750)
z = -0.73

P-value Approach
P-value = 0.4654
As P-value >= 0.1, fail to reject null hypothesis.

There is not sufficient evidence to conclude that the proportion of females who are living alone has changed.