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Slav-nsk [51]
4 years ago
15

In a Harris poll of 514 human resource professionals, 90% said that the appearance of a job applicant is most important for a go

od first impression. Among the 514 human resource professionals who were surveyed, how many of them said that the appearance of a job applicant is most important for a good first impression? Construct a 99% confidence interval estimate of the proportion of all human resource professionals believing that the appearance of a job applicant is most important for a good first impression. Repeat part (b) using a confidence level of 80%. Compare the confidence levels from parts (b) and (c) and identify the interval that is wider. Why is it wider?

Mathematics
1 answer:
jolli1 [7]4 years ago
8 0

Answer:

( a ) 463 " human resource professionals " believe that the appearance of a job applicant is most important for a good first impression

( b )  Percent wise the confidence interval should be from about 86.6% to 93.4%

Step-by-step explanation:

I believe these first two parts here is sufficient enough to get you started!

" In a Harris poll of 514 human resource professionals, 90% said that the appearance of a job applicant is most important for a good first impression. Part A: Among the 514 human resource professionals who were surveyed, how many of them said that the appearance of a job applicant is most important for a good first impression?

Part B: Construct a 99% confidence interval estimate of the proportion of all human resource professionals believing that the appearance of a job applicant is most important for a good first impression. "

_____

( a ) We are given that n = 514, and p = 90%. The number of human resource professionals that said that the appearance of a job applicant is most important for a good first impression, should be the following -

514 x \frac{90}{100}

= \frac{46260}{100}

= 462.6

As the count of people can't be expressed as a fraction, the solution should be about 463 " human resource professionals. "

_____

( b ) Here the confidence level is 0.99 percent. Knowing that -

1 - ∝ = 0.99,

∝ = 1 - 0.99... = 0.01

Therefore, a 99% confidence interval estimate of the proportion of all human resource professionals believing that the appearance of a job applicant is most important for a good first impression should be calculated as directed in the attachment. By that the interval ranges as such -

0.8659 < p < 0.9341

And percent wise the confidence interval should be from about 86.6% to 93.4%

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The sample size needed if the margin of error of the confidence interval is to be about 0.04 is 18.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

Past studies suggest this proportion will be about 0.15

This means that p = 0.15

Find the sample size needed if the margin of error of the confidence interval is to be about 0.04

This is n when M = 0.04. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.04 = 1.96\sqrt{\frac{0.15*0.85}{n}}

0.04\sqrt{n} = 1.96\sqrt{0.15*0.85}

\sqrt{n} = \frac{1.96\sqrt{0.15*0.85}}{0.04}

(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.15*0.85}}{0.04})^{2}

n = 17.5

Rounding up

The sample size needed if the margin of error of the confidence interval is to be about 0.04 is 18.

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