The mean of 15 sample IQ scores of statistic instructors is 118. The sample standard deviation is 11. Given the sample data appear to come from a normally distributed population, find the test statistic for a claim that the population mean IQ of statistic instructors is greater than 110.

A) t = 9.342
B) t = 2.817
C) t = 3.767
D) t = 10.909

1 answer

To determine the test statistic for the claim that the population mean IQ of statistic instructors is greater than 110, we can use the t-test formula for the mean. The formula for the t-statistic is as follows:

\[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \]

Where:

  • \(\bar{x}\) = Sample mean
  • \(\mu_0\) = Population mean under the null hypothesis
  • \(s\) = Sample standard deviation
  • \(n\) = Sample size

Given:

  • Sample mean (\(\bar{x}\)) = 118
  • Hypothesized population mean (\(\mu_0\)) = 110
  • Sample standard deviation (\(s\)) = 11
  • Sample size (\(n\)) = 15

Now we can plug in the values into the formula:

  1. Calculate \(s / \sqrt{n}\):

\[ s / \sqrt{n} = 11 / \sqrt{15} \]

First, calculate \(\sqrt{15}\):

\[ \sqrt{15} \approx 3.872 \]

Now, calculate \(s / \sqrt{n}\):

\[ s / \sqrt{n} \approx 11 / 3.872 \approx 2.840 \]

  1. Plug this value into the t-statistic formula:

\[ t = \frac{118 - 110}{2.840} \approx \frac{8}{2.840} \approx 2.816 \]

Thus, the calculated t-statistic is approximately:

\[ t \approx 2.816 \]

Comparing this result with the provided options:

A) t = 9.342
B) t = 2.817
C) t = 3.767
D) t = 10.909

The closest answer, rounding to three decimal places, suggests that the correct answer is:

B) t = 2.817.