An IQ test is designed so that the mean is 100 and the standard deviation is 1010 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 9595% confidence that the sample mean is within 22 IQ points of the true mean. Assume that sigmaσequals=1010 and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real world calculation.

So the answer for this case would be n=97 rounded up to the nearest integer . And this value is reaonable and possible in the reality.

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".

represent the sample mean for the sample  

population mean (variable of interest)

represent the population standard deviation

n represent the sample size  

Solution to the problem

The margin of error is given by this formula:

   (a)

And on this case we have that ME =2 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

  (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got , replacing into formula (b) we got:

So the answer for this case would be n=97 rounded up to the nearest integer . And this value is reaonable and possible in the reality.

step-by-step explanation:

divide, then simplify your answer.

1. make -15.66 ÷   5.8 a fraction. then divide.

2. simplify.  

3. final answer.  

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