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Out of 2,000 students in a high school, 800 were randomly selected and asked if they want the school's mascot to change. Out of those surveyed, 250 want it to change.

With a desired confidence level of [tex]$99\%$[/tex] and a corresponding [tex]$z^\ \textless \ em\ \textgreater \ $[/tex]-score of 2.58, what is the margin of error for this sample survey?

Complete the statements to find the margin of error:

- The sample size in this problem is [tex]$n = \boxed{800}$[/tex].
- Estimate the population proportion as [tex]$\hat{p} = \boxed{\frac{250}{800}}$[/tex].
- [tex]$(1-\hat{p}) = \boxed{1 - \hat{p}}$[/tex].

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 percent, the result is [tex]$\boxed{E\%}$[/tex].


Sagot :

To find the margin of error for the sample survey, let's break down the solution step by step using the given information.

- The sample size in this problem is [tex]\( 800 \)[/tex].
- The total number of students surveyed who want the mascot to change is [tex]\( 250 \)[/tex].

First, estimate the population proportion, denoted as [tex]\(\hat{p}\)[/tex]:

[tex]\[ \hat{p} = \frac{250}{800} = 0.3125 \][/tex]

Next, calculate [tex]\(1 - \hat{\rho}\)[/tex]:

[tex]\[ 1 - \hat{p} = 1 - 0.3125 = 0.6875 \][/tex]

We are given the [tex]\( z \)[/tex]-score for a [tex]\( 99\% \)[/tex] confidence level, which is [tex]\( 2.58 \)[/tex].

Now, we can calculate the margin of error [tex]\( E \)[/tex] using the formula:

[tex]\[ E = z^* \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \][/tex]

Substitute the values:

[tex]\[ E = 2.58 \cdot \sqrt{\frac{0.3125 \cdot 0.6875}{800}} \][/tex]

Computing this, we get the margin of error. When rounded to the nearest tenth percent, the result is [tex]\( 4.2\% \)[/tex].

Therefore, the complete statements are:
- The sample size in this problem is [tex]\( 800 \)[/tex].
- Estimate the population proportion as [tex]\( 0.3125 \)[/tex].
- [tex]\( (1 - \hat{\rho}) = 0.6875 \)[/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 percent, the result is [tex]\( 4.2 \% \)[/tex].
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