Question

g 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 99​% confidence that the sample mean is within 4 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. The required sample size is nothing. ​(Round up to the nearest​ integer.) Would it be reasonable to sample this number of​ students?

Answer 1

`n=((2.58(18))/(4))^2 =134.79 \approx 135`

So the answer for this case would be n=135 rounded up to the nearest integer

And for this case the value of 135 for a sample is possible and makes sense since is not an impossible value

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`\bar X` represent the sample mean for the sample

`\mu` population mean (variable of interest)

`\sigma` represent the population standard deviation

n represent the sample size

Solution to the problem

Since the Confidence is 0.99 or 99%, the value of
`\alpha=0.01` and
`\alpha/2 =0.005`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that
`z_(\alpha/2)=2.58`

The margin of error is given by this formula:

`ME=z_(\alpha/2)(\sigma)/(√(n))` (a)

And on this case we have that ME =4 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

`n=((z_(\alpha/2) \sigma)/(ME))^2` (b)

Replacing into formula (b) we got:

`n=((2.58(18))/(4))^2 =134.79 \approx 135`

So the answer for this case would be n=135 rounded up to the nearest integer

And for this case the value of 135 for a sample is possible and makes sense since is not an impossible value