Question

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.

Answer #1

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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