Question

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 29?

Answer 1

Using the z-distribution, it is found that she should take a sample of 46 students.

What is a z-distribution confidence interval?

The confidence interval is:

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

The margin of error is:

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

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.

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.

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 found as follows:

$6\sigma = 800 - 200$

$6\sigma = 600$

$\sigma = 100$

The sample size is n when M = 29, hence:

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

$29 = 1.96\frac{100}{\sqrt{n}}$

$29\sqrt{n} = 196$

$\sqrt{n} = \frac{196}{29}$

$(\sqrt{n})^2 = \left(\frac{196}{29}\right)^2$

n = 45.67.

Rounding up, a sample of 46 students should be taken.

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

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