Question

question 1. A random sample of 1002 adults in a certain large country was asked " do you pretty much think televisions are necessity or a luxury you could do without?" Of the 1002 adults surveyed, 511 indicated that televisions are a luxury they could do without. Complete parts (a) through (e) below

question 2.

A researcher wishes to estimate the proportion of adults who have​ high-speed Internet access. What size sample should be obtained if she wishes the estimate to be within

0.030.03

with

9595​%

confidence if

​(a) she uses a previous estimate of

0.380.38​?

​(b) she does not use any prior​ estimates?

question 3.

A simple random sample of size n is drawn from a population that is normally distributed. The sample​ mean,

x overbarx​,

is found to be

111111​,

and the sample standard​ deviation, s, is found to be

1010.

​(a) Construct

aa

9090​%

confidence interval about

muμ

if the sample​ size, n, is

1919.

​(b) Construct

aa

9090​%

confidence interval about

muμ

if the sample​ size, n, is

1313.

​(c) Construct

anan

8080​%

confidence interval about

muμ

if the sample​ size, n, is

1919.

​(d) Could we have computed the confidence intervals in parts​ (a)-(c) if the population had not been normally​ distributed?

question 4.

The following data represent the pH of rain for a random sample of 12 rain dates. A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot indicates there are no outliers. Complete parts​ a) through​ d) below. question 5. People were polled on how many books they read the previous year. Initial survey results indicate that sequals=13.813.8 books. Complete parts ​(a) through ​(d) below. (a) How many subjects are needed to estimate the mean number of books read the previous year within 95% confidence?

5.055.05 5.72 4.624.62 4.80 5.02 4.604.60 4.74 5.19 4.614.61 4.76 4.56 5.685.68

Answer 1

A researcher wishes to estimate the proportion of adults who have​ high-speed Internet access. What size sample should be obtained if she wishes the estimate to be within

0.03

with

95​%

confidence if

​(a) she uses a previous estimate of

0.38

Z crit for 95%=1.96

p=0.38

E=0.03

we have n=(zcrit/E)^2*p*(1-p)

=(1.96/0.03)^2*0.38*(1-0.38)

=1005.645

n=1006

required sample size=1006

​(b) she does not use any prior​ estimates?

so p=0.5

we have

=(zcrit/E)^2*p*(1-p)

=(1.96/0.03)^2*0.5*(1-0.5)

=1067.111

n=1067

ANSWER:

required sample size=1067