Question

in a poll of 517 human resource professionals, 45.6% said that body piercings and tattoos were big personal grooming red flags. complete parts (a) through (d) below.
a. among the 517 human resource professionals who were surveyed, how many of them said that body piercings and tattoos were big personal grooming red flags?
236 (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 personal grooming red flags.
.400 < p <.512
(round to three decimal places as needed.)
c. repeat part (b) using a confidence level of 80%.
<p<
(round to three decimal places as needed.)

Explanation:

Step1: Identify the formula for confidence interval

The formula for a confidence interval for a proportion is $\hat{p}\pm z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$, where $\hat{p}$ is the sample proportion, $n$ is the sample size, and $z$ is the z - score corresponding to the confidence level.

Given $\hat{p}=0.456$, $n = 517$.

For an $80\%$ confidence level, the significance level $\alpha=1 - 0.80=0.20$, and $\alpha/2=0.10$.

From the standard normal distribution table, $z_{\alpha/2}=z_{0.10}\approx1.28$.

Step2: Calculate the margin of error

First, calculate $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$:

$\hat{p}(1 - \hat{p})=0.456\times(1 - 0.456)=0.456\times0.544 = 0.248064$

$\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{\frac{0.248064}{517}}\approx\sqrt{0.0004798}\approx0.0219$

The margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=1.28\times0.0219\approx0.0280$

Step3: Calculate the confidence interval

The lower limit is $\hat{p}-E=0.456 - 0.0280=0.428$

The upper limit is $\hat{p}+E=0.456+0.0280 = 0.484$

Answer:

$0.428 < p < 0.484$