Question

the standard iq test is designed so that the mean is 100 and the standard deviation is 15 for the population of all adults. We wish to find the sample size necessary to estimate the mean IQ score of statistics students. Suppose we want to be 90% confident that our sample mean is within 2 IQ points of the true mean. Assume then that

Answer 1

Answer:

A sample of 153 is needed.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$.

That is z with a pvalue of $1 - 0.05 = 0.95$, so Z = 1.645.

Now, find the margin of error M as such

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

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

Standard deviation is 15 for the population of all adults.

This means that $\sigma = 15$

We wish to find the sample size necessary to estimate the mean IQ score of statistics students. Suppose we want to be 90% confident that our sample mean is within 2 IQ points of the true mean.

This means that a sample of n is needed.

n is found when M = 2. So

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

$2 = 1.645\frac{15}{\sqrt{n}}$

$2\sqrt{n} = 1.645*15$

Simplifying both sides by 2

$\sqrt{n} = 1.645*7.5$

$(\sqrt{n})^2 = (1.645*7.5)^2$

$n = 152.2$

Rounding up:

A sample of 153 is needed.