Question

The admissions director at big city university proposed using the iq scores of current students as a marketing tool. the university agrees to provide him with enough money to administer the iq tests to 50 students. the director gives the test to a simple random sample of 50 of the university's 5000 freshman. the mean iq score for the sample is the iq test that he administered is known to have a standard deviation of what is the 95% confidence interval for the sample mean ? that is what can the director say about the mean score of the population of all 5000 freshman?

Answer 1

The complete question is as follows:

The admission directory of Big City University has a novel idea. He proposed using the IQ scores of current students as a marketing tool. The university agrees to provide him with enough money to administer IQ tests to 50 students. So the director gives the IQ test to an SRS of 50 of the university’s 5000 freshman. The mean IQ score for the sample is xbar=112. The IQ test he administered is known to have a σ of 15. What is the 95% Confidence Interval about the mean? What can the director say about the mean score of the population of all 5000 freshman?

Answer: The 95% confidence interval about the mean is $Confidence interval = 107.84 \leq \mu \leq 116.16$.

The director can say that he is 95% confident that the mean IQ score of the 5000 freshmen lies between 107.84 and 116.16.

We follow these steps to arrive at the answer:

Since the population standard deviation of the IQ test is known, we can use the Z scores to find the confidence interval.

The formula for the confidence interval about the mean is:

$Confidence interval = \overline{X}\pm Z*\frac{\sigma}{\sqrt{n}}$

In the equation above, X bar is known as the point estimate and the second term is known as Margin of Error.

The Critical Value of Z at the 95% confidence level is 1.96.

Substituting the values in the question in the equation above we have,

$Confidence interval = \112\pm 1.96*\frac{15}{\sqrt{50}}$

$Confidence interval = \112\pm 4.157787873}$

$Confidence interval = 107.8422121 \leq \mu \leq 116.1577879$