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In a poll of 512 human resource professionals, [tex]45.7 \%[/tex] said that body piercings and tattoos were big personal grooming red flags. Complete parts (a) through (d) below.

a. Among the 512 human resource professionals who were surveyed, how many of them said that body piercings and tattoos were big personal grooming red flags?
[tex]\[ 234 \][/tex]
(Round to the nearest integer as needed.)

b. Construct a [tex]99 \%[/tex] confidence interval estimate of the proportion of all human resource professionals believing that body piercings and tattoos are big personal grooming red flags.
[tex]\[ 0.400 \ \textless \ p \ \textless \ 0.514 \][/tex]
(Round to three decimal places as needed.)

c. Repeat part (b) using a confidence level of [tex]80 \%[/tex].
[tex]\[ \square \ \textless \ p \ \textless \ \square \][/tex]
(Round to three decimal places as needed.)



Answer :

GinnyAnswer
Let's go through each part of the question step-by-step.

### Part (a)
Among the 512 human resource professionals who were surveyed, 45.7% said that body piercings and tattoos were big personal grooming red flags.

To find the number of HR professionals who said this, we can use the proportion and the total number of surveyed professionals:
[tex]\[ \text{Number of HR professionals} = 512 \times 0.457 \][/tex]

Rounding this to the nearest integer:
[tex]\[ \text{Number of HR professionals} \approx 234 \][/tex]

So, 234 HR professionals out of 512 said that body piercings and tattoos were big personal grooming red flags.

### Part (b)
We need to construct a 99% confidence interval for the proportion of HR professionals who believe body piercings and tattoos are big personal grooming red flags.

Given:
- Sample proportion ([tex]\( \hat{p} \)[/tex]) = 0.457
- Number of surveyed HR professionals (n) = 512

For a 99% confidence interval, we need to find the margin of error [tex]\( E \)[/tex]:
[tex]\[ E = Z \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \][/tex]

Using the z-score for a 99% confidence level, which is approximately 2.576, we find the margin of error and then construct the confidence interval:
[tex]\[ CI_{99\%} = \left( \hat{p} - E, \hat{p} + E \right) \][/tex]

This gives us the confidence interval:
[tex]\[ 0.400 < p < 0.514 \][/tex]

### Part (c)
We need to construct an 80% confidence interval for the proportion of HR professionals who believe body piercings and tattoos are big personal grooming red flags.

Given:
- Sample proportion ([tex]\( \hat{p} \)[/tex]) = 0.457
- Number of surveyed HR professionals (n) = 512

For an 80% confidence interval, the z-score value is approximately 1.282. We use the formula for the margin of error:
[tex]\[ E = Z \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \][/tex]

Using the z-score for an 80% confidence level, and then constructing the confidence interval:
[tex]\[ CI_{80\%} = \left( \hat{p} - E, \hat{p} + E \right) \][/tex]

This gives us the confidence interval:
[tex]\[ 0.429 < p < 0.485 \][/tex]

So, summarizing:
- (a) Number of HR professionals who said body piercings and tattoos were big personal grooming red flags: 234
- (b) 99% confidence interval for the proportion: [tex]\( 0.400 < p < 0.514 \)[/tex]
- (c) 80% confidence interval for the proportion: [tex]\( 0.429 < p < 0.485 \)[/tex]

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