One of the relatively uncommon instances in which researchers know the population standard deviation is in the case of the Intelligence Quotient or IQ test. In general, the average IQ score in large, diverse populations is 100 and the standard deviation is 15. Suppose that your sample of 300 members of your community gives you a mean IQ score of 108. Calculate a 90% confidence interval for the mean and indicate which answer set comes closest to those that would fill the blanks in the following interpretation: we can be 90% confident that the mean IQ score in this community lies between and .

90% confidence interval for the mean IQ score in this community is [106.58 , 109.42].

Step-by-step explanation:

We are given that in general, the average IQ score in large, diverse populations is 100 and the standard deviation is 15.

Suppose that your sample of 300 members of your community gives you a mean IQ score of 108.

Firstly, the pivotal quantity for 90% confidence interval for the population mean is given by;

                        P.Q. =  ~ N(0,1)

where, = sample mean IQ score = 108

             = population standard deviation = 15

            n = sample of members = 300

             = population mean IQ score

Here for constructing 90% confidence interval we have used One-sample z test statistics because we know about population standard deviation.

So, 90% confidence interval for the population mean, is ;

P(-1.645 < N(0,1) < 1.645) = 0.90  {As the critical value of z at 5% level

                                                of significance are -1.645 & 1.645}  

P(-1.645 < < 1.645) = 0.90

P( < < ) = 0.90

P( < < ) = 0.90

90% confidence interval for = [ , ]

                                      = [ , ]

                                      = [106.58 , 109.42]

Therefore, we can be 90% confident that the mean IQ score in this community lies between 106.58 and 109.42.

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step-by-step explanation:

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