Question

(31 points) Wilson County School District consists of 2,548 students. The district decided to conduct a survey regarding their new dress code policy. Wilson County School District surveyed 479 of their students and found that 42% of those surveyed disliked the new dress code policy. What is the margin of error, assuming a 95% confidence level?

Answer 1

Answer:

+ or - 0.044

Step-by-step explanation:

Answer 2

The margin of error is 0.044.

What is a Margin of Error?

A Margin of Error is defined as the results will vary from the actual population value by how many percentage points, as indicated by the margin of error.

The margin of error using this formula, where z is the critical value.

$ME = z\sqrt{\dfrac{T(1 - t)}{n} }$

To determine a margin of error for a sample proportion,

Where n is the sample size and is the sample proportion, these assumptions must be true.

n t ≥ 10

n(1 - t) ≥ 10

Write the proportion of lead to significant as a decimal number first,

42%

= 0.42

Now substitute the sample proportion and the sample size, 479, the assumptions are inserted into the equations to see if they are satisfied.

⇒(479X0.42) 201.18

⇒ (479)(1 − 0.42)

⇒  277.82

Both values are greater than 10, so the assumptions are satisfied, and the margin of error can be calculated.

Determine the margin of error using this formula, where z is the critical value.

$ME = z\sqrt{\dfrac{T(1 - t)}{n} }$

For a 95% confidence level, z = 1.96.

Substitute the values of z, n, and T into the formula for margin of error and evaluate.

$ME = 1.96\sqrt{\dfrac{0.42(1 - (10.42)}{479} }$

ME ≈ 0.044

Hence, the margin of error is 0.044.

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