A report on electronic monitoring and surveillance summarized a survey of 306 U.

Question

A report on electronic monitoring and surveillance summarized a survey of 306 U.S. businesses. The report stated that 93 of the 306 businesses had fired workers for misuse of the Internet. Assume that this sample is representative of businesses in the United States. (a) Estimate the proportion of all businesses in the U.S. that have fired workers for misuse of the Internet. (Round your answer to three decimal places.) What statistic did you use? sample size sample variance sample proportion sample mean sample standard deviation (b) Use the sample data to estimate the standard error of p. (Round your answer to three decimal places.) (c) Calculate the margin of error associated with the estimate in part (a). (Hint: See Example 9.3. Round your answer to three decimal places.) Interpret the margin of error associated with the estimate in part (a). (Hint: See Example 9.3.) It is unlikely that the estimate in part (a) differs from the true population proportion by less than the value calculated above. It is unlikely that the estimate in part (a) differs from the true population proportion by more than the value calculated above. It is impossible that the estimate in part (a) differs from the true population proportion by more than the value calculated above It is impossible that the estimate in part (a) differs from the true population proportion by less than the value calculated above. It is likely that the estimate in part (a) differs from the true population proportion by the value calculated above

Explanation / Answer

a)
p = 93/306 = 0.30

Sample proportion

b)

Standard error = sqrt( p * (1-p)/n)
= sqrt( 0.3 * (1-0.3) /306)
= 0.03
c)
z value at 95% = 1.96

Margin of error = z * sqrt( p * (1-p)/n)
= 1.96 * sqrt( 0.3 * (1-0.3) /306)
= 0.051
It is unlikely that the estimate in part a) differes from the true population proportion by more than the value calculated above

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