oblem 8-1: Black and White and Read All Over: Reading Habits of Adult Males In a

Question

oblem 8-1: Black and White and Read All Over: Reading Habits of Adult Males In a Midwestern city, a simple random sample of75 adult men was taken. Each man was asked how much time he spent reading books per week. The results of the sample showed a mean of 73 minutes with a standard deviation of 21 minutes. Construct and interpret the 9o% confidence interval of the mean for all males in this city. Problem #8-2: After School Activities (Margin of error with proportions) A simple random sample of 10o FCPS students indicated that 83% of students are involved in at least one after-school activity. Construct and interpret the 88% confidence interval of the proportion of students who are involved in at least one after-school activity. Problem #8-3: Education choice A national opinion poll is being designed to determine what percentage of American adults agree that parents should be given vouchers that are good for education at any public or private school of their choice. How large of a simple random sample is required in order to obtain a margin of error of +/-3% in a 99% confidence interval?

Explanation / Answer

Question 1

n=sample size=75

X bar= sample mean= 73

s= sample sd=21

Confidence Interval= Xbar+/- z(SE)

z value at 90% CI=+/-1.64

Putting the values above we get

73 +/- 1.64(21/8.66)

as SE= Sample sd/ rt n

thus, CI=76.3976, 69.0231

It implies the accepatable range within which the average reading time can fluctuate

Question 2

Percentage of the students involved in atleast one after school activity is p= 0.83

the CI at 88%

CI= 0.83+/- z(SE)

    =0.83+/- 1.56(0.3/sample sd/rt 10)

since z value at 88 % CI=1.56 and rt 100=10

thus, CI= 0.83+/-4.86/sample sd

Question 3

Margin of error=ME=p-P=+/-0.03

p= sample proportion of the American adults who agree that vouchers should be given to parents

P= Actual proportion of the American adults who agree that vouchers should be given to parents

At 99% CI the z-value equals 2.82

and SE= ME/ SD/rt n

CI=Confidence Interval= p+/- z(SE)

Thus,

CI=0.83+/-2.82(0.03rt n/sd)

now from

ME= p-P/sd/rtn

+/-0.03= p-P/sd/rtn

bring rt n to one side and we get

n=(0.036/p-P)2

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