Question

An IQ test is designed so that the mean is 100 and the standard deviation is 18 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of statistics students such that it can be said with 90​% confidence that the sample mean is within 3 IQ points of the true mean. Assume that sigmaequals18 and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real world calculation.

Answer 1

Answer:

At least 98 people need to be sampled. It is not a huge number, that is, it is not difficult to sample 98 people, so it is a reasonable sample size for a real world calculation.

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 $1-\alpha$.

So it 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.

In this question:

We need a sample size of at least n.

n is found when M = 3. We have that $\sigma = 18$.

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

$3 = 1.645*\frac{18}{\sqrt{n}}$

$3\sqrt{n} = 1.645*18$

Simplifying by 3

$\sqrt{n} = 1.645*6$

$(\sqrt{n})^{2} = (1.645*6)^{2}$

$n = 97.41$

Rounding up,

At least 98 people need to be sampled. It is not a huge number, that is, it is not difficult to sample 98 people, so it is a reasonable sample size for a real world calculation.