Question

In a poll of 506 human resource​ professionals, 48.2% said that body piercings and tattoos were...

In a poll of 506 human resource​ professionals, 48.2%

said that body piercings and tattoos were big grooming red flags. Complete parts​ (a) through​ (d) below.

​a) Among the 506 human resource professionals who were​ surveyed, how many of them said that body piercings and tattoos were big grooming red​ flags?

​b) Construct a​ 99% confidence interval estimate of the proportion of all human resource professionals believing that body piercings and tattoos are big grooming red flags. nothingless than p less than<p<nothing

​(Round to three decimal places as​ needed.)

​c) Repeat part​ (b) using a confidence level of​ 80%. nothingless than p less than<p<nothing

​(Round to three decimal places as​ needed.)

​d) Compare the confidence intervals from parts​ (b) and​ (c) and identify the interval that is wider. Why is it​ wider?

Select the correct choice below and fill in the answer boxes to complete your choice.

A. The nothing​ _% confidence interval is wider than the nothing​ _% confidence interval. As the confidence interval​ narrows, the probability that the confidence interval actually does contain the population parameter increases.

B. The nothing​ _% confidence interval is wider than the nothing​ _% confidence interval. As the confidence interval​ widens, the probability that the confidence interval actually does contain the sample parameter increases.

C.The nothing​ _% confidence interval is wider than the nothing​ _% confidence interval. As the confidence interval​ widens, the probability that the confidence interval actually does contain the population parameter increases.

D. The nothing​ _% confidence interval is wider than the nothing​ _% confidence interval. As the confidence interval​ narrows, the probability that the confidence interval actually does contain the sample parameter increases.

a )

Expected value = n * p

= 506 * 0.482

= 243.892 ( 244)

b)

99% confidence interval for proportion p is

- Z/2 * Sqrt( ( 1 - ) / n) < p < + Z/2 * Sqrt( ( 1 - ) / n)

0.482 - 2.576 * sqrt( 0.482 * 0.518 / 506) < p < 0.482 + 2.576 * sqrt( 0.482 * 0.518 / 506)

0.425 < p < 0.539

c)

80% confidence interval for proportion p is

- Z/2 * Sqrt( ( 1 - ) / n) < p < + Z/2 * Sqrt( ( 1 - ) / n)

0.482 - 1.2816 * sqrt( 0.482 * 0.518 / 506) < p < 0.482 + 1.2816 * sqrt( 0.482 * 0.518 / 506)

0.454 < p < 0.510

d)

The 99% confidence interval is wider than 80% confidence interval. As the cnfidence interval widens,

the probability that the confidence interval actually does contain the population parameter increases.

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