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A recent poll of 500 employees from a company of 1,300 employees was conducted to see how many of them believe the minimum wage should be raised. Of those polled, 435 feel that the minimum wage should be raised.

With a desired confidence level of [tex]$95\%$[/tex], and a corresponding [tex]$z^\ \textless \ em\ \textgreater \ $[/tex]-score of 1.96, what is the margin of error for this sample survey?

Complete the statements to find the margin of error:
1. The sample size in this problem is [tex]\square[/tex].
2. The population proportion is estimated as [tex]\square[/tex].
3. When the margin of error is calculated using the formula [tex]$E=z^\ \textless \ /em\ \textgreater \ \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$[/tex], to the nearest tenth of a percent, the result is [tex]\square[/tex].



Answer :

GinnyAnswer
Let's complete the statements step-by-step:

1. Identify the sample size:
The sample size is the number of employees polled.
- The sample size in this problem is [tex]\(500\)[/tex] employees.

2. Calculate the population proportion:
The population proportion is the fraction of the sample who support raising the minimum wage.
- The population proportion is estimated as [tex]\(\frac{435}{500} = 0.87\)[/tex].

3. Determine the [tex]\(z^\)[/tex]-score:
The [tex]\(z^\)[/tex]-score for a 95% confidence level is given as 1.96.

4. Calculate the margin of error using the formula:
The formula for the margin of error [tex]\(E\)[/tex] is:
[tex]\[ E = z^* \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \][/tex]
where:
- [tex]\(\hat{p}\)[/tex] is the population proportion (0.87),
- [tex]\(n\)[/tex] is the sample size (500),
- and [tex]\(z^*\)[/tex] is the z-score (1.96).

5. Substitute the values into the formula:
[tex]\[ E = 1.96 \cdot \sqrt{\frac{0.87 \cdot (1 - 0.87)}{500}} \][/tex]

6. Simplify the expression:
[tex]\[ E = 1.96 \cdot \sqrt{\frac{0.87 \cdot 0.13}{500}} \][/tex]
[tex]\[ E = 1.96 \cdot \sqrt{\frac{0.1131}{500}} \][/tex]
[tex]\[ E = 1.96 \cdot \sqrt{0.0002262} \][/tex]
[tex]\[ E = 1.96 \cdot 0.01504 \approx 0.0295 \][/tex]

7. Convert to a percentage and round to the nearest tenth:
To express the margin of error as a percentage, multiply by 100.
[tex]\[ 0.0295 \times 100 = 2.95\% \][/tex]
Rounding to the nearest tenth of a percent gives us [tex]\(2.9\%\)[/tex].

Therefore, the completed statements are:

- The sample size in this problem is [tex]\(500\)[/tex] employees.
- The population proportion is estimated as [tex]\(0.87\)[/tex].
- When the margin of error is calculated using the formula [tex]\(E = z^* \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\)[/tex], to the nearest tenth of a percent, the result is [tex]\(2.9\%\)[/tex].

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