Question

3.

a. A survey of 1000 U.S. adults found that 34% of people said that they would get no work done on Cyber Monday since they would spend all day shopping online. Find the 90% confidence interval of the true proportion. ROUND TO FIVE DECIMAL PLACES

b. In a recent year, 6% of cars sold had a manual transmission. A random sample of college students who owned cars revealed the following: out of 129 cars, 23 had manual transmissions. Estimate the proportion of college students who drive cars with manual transmissions with 99% confidence. ROUND TO THREE DECIMAL PLACES

c. A random sample of 325 college students were asked if they believed that places could be haunted, and 147 responded yes. Estimate the true proportion of college students who believe in the possibility of haunted places with 95% confidence. According toTime magazine, 37% of Americans believe that places can be haunted. ROUND TO THREE DECIMAL PLACES

___ <p<____

d. A CBS News/New York times poll found that 329 out of 763 randomly selected adults said they would travel to outer space in their lifetime, given the chance. Estimate the true proportion of adults who would like to travel to outer space with 94% accuracy. ROUND TO THREE DECIMAL PLACES

___ <p < ____

e. Took a random sample of 200 people, 142 said that they watched educational television. Find the 95% confidence interval of the true proportion of people who watched educational television. ROUND TO THREE DECIMAL PLACES

____ < p < ____

Answer 1

Solution_A:

n=1000

p^=-0.34

z crit for 90%=1.645

90% confidence interval for p

p^-Z*sqrt(p^*(1-p^)/n),p^+Z*sqrt(p^*(1-p^)/n)

0.34-1.645*sqrt(0.34*(1-0.34)/1000),0.34+1.645*sqrt(0.34*(1-0.34)/1000)

0.3153579,0.3646421

(0.31536,0.36464)

sOLUTION-B:

Sample proportion of manual transmission,p^=x/n=23/129= 0.1782946

z crit for 99%=2.576

99% confidence interval for p is

p^-Z*sqrt(p^*(1-p^)/n),p^+Z*sqrt(p^*(1-p^)/n)

0.1782946-2.576*sqrt(0.1782946*(1-0.1782946)/129),0.1782946+2.576*sqrt(0.1782946*(1-0.1782946)/129)

0.09148292,0.2651063

0.092,0.265

ANSWER"

99% confidene interval

=0.092<p<0.265

Solution-b:

z crit for 95%=1.96

sampl proportion proportion of college students who believe in the possibility of haunted places

p^=x/n=147/325

= 0.4523077

95% confidence interval for p is

p^-Z*sqrt(p^*(1-p^)/n),p^+Z*sqrt(p^*(1-p^)/n)

0.4523077-1.96*sqrt(0.4523077*(1-0.4523077)/325),0.4523077+1.96*sqrt(0.4523077*(1-0.4523077)/325)

0.3981949, 0.5064205

0.398,0.506

95% confidence interval for p is

0.398<p<0.506

Solution-d:

z crit for 94%=NORM.S,INV(0.97)=1.880793608

p^=x/n=329/763= 0.4311927

94% confidence interval for p is

0.4311927-1.880793608*sqrt(0.4311927*(1-0.4311927)/763),0.4311927+1.880793608*sqrt(0.4311927*(1-0.4311927)/763)

0.397472,0.4649134

0.397<p<0.465