Question

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

Answer 1

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 \(s / \sqrt{n}\):

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

2. 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**.