Question

A recent 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 970 subjects, 25 answered yes and 945 answered no. (a) What is the statistic used to estimate the proportion of the population who were victims. (b) Find the standard error of this estimate. (3 decimal places) (c) Find the margin of error for a 99% confidence interval. (3 decimal places) d) Given the 99% confidence interval or the population proportion is 0.013 and 0.039, can you conclude that er er than 10% of all ad s n the United States erent ms No, the value is not included in the interval Yes, the value is included in the interval. Yes, all of the values in the interval are below 10%. No, all of the values in the interval are below 10%. No conclusion can be drawn from the interval

Answer 1

Here' the answer to the question with full concept. Please don't hesitate to give a "thumbs up" in case you're satisfied with the answer

a.

Statistic used to estimate the proportion of the population who are victims

= 25/970

= 0.026

b.

Standard error, SE

= sqrt(p*p'/n)

= sqrt(.026*(1-.026)/970)

= 0.005

c.

Margin of error for a 99% CI has a Z value of 2.575

So, Margin of error = Z*SE = 2.575*.005 = 0.013

d.

So, actual 99% CI is .026+/-.013 or .013 to .039 is the 99% CI.

So, to have fewer than 10% of all adults in US were victims is not usual because values less than .10 lie in the range .013 to .039

Hence, Yes, all the values in the interval are below 10%

So, Option C (3rd) is correct