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Crime victims In 1994 (the most recent year asked), the General Social Survey asked, “During the last year, did anyone take something from you by using force—such as a stickup, mugging, or threat?” Of 1223 subjects, 31answered yes and 1192 answered no.
a. Find the point estimate of the proportion of the population who were victims.
b. Find the standard error of this estimate.
c. Find the margin of error for a 95% confidence interval.
d. Construct the 95% confidence interval for the population proportion. Can you conclude that fewer than
10% of all adults in the United States were victims?

Sagot :

fichoh

Answer:

0.0253475; 0.0044944 ; 0.08809% ;(0.01654, 0.03416) ;

Explanation:

Given that:

n = 1223 ;

Yes = 31 ; No = 1192

a. Find the point estimate of the proportion of the population who were victims.

Yes / n = 31 / 1223 = 0.0253475

b. Find the standard error of this estimate.

Standard Error (se) :

Sqrt[(p(1 - p)) / n]

S. E= sqrt[(0.0253475(1 - 0.0253475)) / 1223]

S. E = √2.0200 * 10^-5

S.E = 0.0044944

c. Find the margin of error for a 95% confidence interval.

MOE = Zcritical * S. E

Zcritical at 95% = 1.96

MOE = 1.96 * 0.0044944

MOE = 0.008809024

MOE = 0.008809024 * 100%

MOE = 0.08809%

d. Construct the 95% confidence interval for the population proportion. Can you conclude that fewer than

10% of all adults in the United States were victims?

Confidence interval :

Mean ± MOE

0.0253475 ± 0.008809024

LOWER BOUNDARY :

0.0253475 - 0.008809024 = 0.016538476

UPPER BOUNDARY :

0.0253475 + 0.008809024 = 0.034156524

(0.01654, 0.03416)

Fewer than 10% of alduts are victims (10% = 0.1) ; interval values is less than 0.1

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