Mathematics, 11.03.2020 18:01 queenfan1818

An IQ test is designed so that the mean is 100 and the standard deviation is 2222 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 9090% confidence that the sample mean is within 44 IQ points of the true mean. Assume that sigmaσequals=2222 and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real world calculation.

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Answer from: ashleyzamarripa08

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (2)

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 (2) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (3)

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ 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 $z_{\alpha/2}=1.64$

$n=(\frac{1.64(22)}{4})^2 = 81.36\approx 82$

So the answer for this case would be n=82 rounded up to the nearest integer

Step-by-step 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

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (2)

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 (2) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (3)

$n=(\frac{1.64(22)}{4})^2 = 81.36\approx 82$

So the answer for this case would be n=82 rounded up to the nearest integer

Answer from: Quest

It’s hard to understand what you put.. can you show a picture ?

Answer from: Quest

answer are

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An IQ test is designed so that the mean is 100 and the standard deviation is 2222 for the population...

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