Question

A random sample of 1003 adult Americans was asked, "Do you pretty much think televisions are a necessity or a luxury you could do without?" Of the 1003 adults surveyed, 521 indicated that televisions are a luxury they could do without.
Requried:
a. Obtain a point estimate for the population proportion of adult Americans who believe that televisions are a luxury they could do without.
b. Verify that the requirements for constructing a confidence interval about p are satisfied.
c. Construct and interpret a 95% confidence interval for the population proportion of adult Americans who believe that televisions are a luxury they could do without.
d. Is it possible that a supermajority (more than 60%) of adult Americans believe that television is a luxury they could do without ? Is it likely?
e. Use the results of part (c) to construct a 95% confidence interval for the population proportion of adult Americans who believe that televisions are a necessity.

Answer 1

Answer:

a)0.519

b)requirements for constructing a confidence interval about p are satisfied.

c)0.488, and 0.550

d)yes this is because there is no possibility

that the true proportion is not captured in the confidence interval.

e)

0.450, 0.512

Step-by-step explanation:

a)The point estimate for the population proportion of adult Americans who believe that televisions are a luxury they could do without can be calculated as

521 of that adult that indicated that televisions are a luxury they could do without/1003 adults surveyed,

p = 521/1003

= 0.519.

b)

np(1-p)=1003×(0.519)×(1-0.519)

=250.39≥10 and the sample size is less than 5% of the population

therefore, requirements for constructing a confidence interval about p are satisfied.

c) we were given 95% confidence,

Then α=(1-0.95)

= 0.05

From the Z-tables,we can get the critical value is Z , which is (0.05/2) = Z (0.025) = 1.96

confidence interval can the be calculated using the formula below

- p ± Z*√ p (1 – p)/n

=0.519 ± 1.96√0.519×(1 – 0.519)/(1003)

= 0.519 ± 1.96×0.0158

0.519 ± 0.031

= 0.488, and 0.550

d) yes this is because there is no possibility

that the true proportion is not captured in the confidence interval.

e)the sample size can be calculated as

P=x/n

=(1003-521)/1003

=0.481

But we're given 95% confidence interval

Then

α=(1-0.95)=

0.05

From the Z-tables,we can get the critical value is Z , which is (0.05/2) = Z (0.025) = 1.96

Then convidence interval=

- p ± Z*√ p (1 – p)/n

=0.481 ± 1.96√0.481×(1 – 0.481)/(1003)

0.450, 0.512