In a poll of 542 human resource professionals 544 said that

In a poll of

542

human resource professionals,

54.4

%

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

a) Among the

542

human resource professionals who were surveyed, how many of them said that body piercings and tattoos were big grooming red flags?

295

(Round to the nearest integer as needed.)

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.

nothing

less than p less thannothing

(Round to three decimal places as needed.)

c) Repeat part (b) using a confidence level of 80%.

nothing

less than p less thannothing

(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 widens, 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 narrows, the probability that the confidence interval actually does contain the population parameter increases.

C.

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.

D.

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.

Click to select your answer(s).

Solution

a) out of 542 persons 54.4% said that body piercings and tattoos were big grooming red flags.

hence the number of people=542*54.4/100=294.848=295 [answer]

b) the estimate of the proporton be b=295/542=0.544

then a 99% confidence interval of p is

[b-sqrt(b(1-b)/542)*tao_0.005,b+sqrt(b(1-b)/542)*tao_0.005]

where tao_0.005 is the upper 0.005 point of a normal(0,1) distribution

tao_0.005=2.57583

hence the confidene interval is

[0.488<p<0.599] [answer]

c) similarly 80% confidence interval is

[b-sqrt(b(1-b)/542)*tao_0.1,b+sqrt(b(1-b)/542)*tao_0.1]

where tao_0.1 is the upper 0.1 point of a normal(0,1) distribution

tao_0.1=1.28155

hence the confidene interval is

[0.517<p<0.571] [answer]

d) the width of the interval in b) is [0.599-0.488]=0.111

the width of the interval in c) is [0.571-0.517]=0.054

hence the first interval is wider

hence the correct alternative is the 99% confidence interval is wider than the 80% confidence interval.As the confidence interval widens, the probability that the confidence interval actually does contain the sample parameter increases. [answer]