Question

The SAT is the most widely used test in the undergraduaté admissions process, Scores on the math portion of the SAT are believed to
be normally distributed and range from 200 to 800. A researcher from the admissions department at the University of New Hampshire
is interested in estimating the mean math SAT scores of the incoming class with 95% confidence. How large a sample should she take
to ensure that the margin of error is below 20?
(You may find it useful to reference the z table. Round intermediate calculations to at
least 4 decimal places and "2" value to 3 decimal places. Round up your final answer to the nearest whole number.)

Answer 1

Using the z-distribution, it is found that she should take a sample of 97 scores to ensure that the margin of error is below 20.

What is a z-distribution confidence interval?

The confidence interval is:

$\overline{x} \pm z\frac{\sigma}{\sqrt{n}}$

In which:

• $\overline{x}$ is the sample mean.

• z is the critical value.

• n is the sample size.

• $\sigma$ is the standard deviation for the population.

The margin of error is given by:

$M = z\frac{\sigma}{\sqrt{n}}$

Scores on the math portion of the SAT are believed to be normally distributed and range from 200 to 800, hence by the Empirical Rule the standard deviation is given by:

$6\sigma = 800 - 200$

$\sigma = 100$

In this problem, we have a 95% confidence level, hence$\alpha = 0.95$, z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$, so the critical value is z = 1.96.

To ensure a margin of error of 20, we solve for n when M = 20, hence:

$M = z\frac{\sigma}{\sqrt{n}}$

$20 = 1.96\frac{100}{\sqrt{n}}$

$20\sqrt{n} = 1.96 \times 100$

$\sqrt{n} = 1.96 \times 5$

$(\sqrt{n})^2 = (1.96 \times 5)^2$

n = 96.04.

Rounding up, she should take a sample of 97 scores to ensure that the margin of error is below 20.

More can be learned about the z-distribution at brainly.com/question/25890103

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