Use your TI83: A claim is made that the average human IQ is 104. A sample of 37 such IQs yields a sample average of 106. If the population deviation is 12, determine the p-value (nearest thousandth) to test that the population average IQ is not 104.

To determine the p-value for testing the claim that the population average IQ is not 104, we will perform a hypothesis test using the information given. Here's how you can do it step-by-step:

State the Hypotheses:
- Null Hypothesis (H₀): [tex]\mu = 104[/tex] (The population mean IQ is 104)
- Alternative Hypothesis (H₁): [tex]\mu \neq 104[/tex] (The population mean IQ is not 104)
Identify the Test Statistic:
We will use the z-test because we know the population standard deviation. The test statistic is calculated using the formula:
[tex]z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
where:
- [tex]\bar{x} = 106[/tex] (sample mean)
- [tex]\mu = 104[/tex] (hypothesized population mean)
- [tex]\sigma = 12[/tex] (population standard deviation)
- [tex]n = 37[/tex] (sample size)
Calculate the Z-score:
[tex]z = \frac{106 - 104}{\frac{12}{\sqrt{37}}}[/tex]
[tex]z \approx \frac{2}{1.973} \approx 1.013[/tex]
Determine the P-value:
Since this is a two-tailed test, we need to find the probability that z is at least 1.013 or at most -1.013.
Using a standard normal distribution table or calculator, find the probability associated with [tex]z = 1.013[/tex].
[tex]P(Z \geq 1.013) \approx 0.155[/tex]
Since it's two-tailed, double this probability:
[tex]P-value = 2 \times 0.155 \approx 0.310[/tex]
Conclusion:
Compare the p-value with the significance level (typically [tex]\alpha = 0.05[/tex]). If the p-value is less than [tex]\alpha[/tex], reject the null hypothesis.
In this case, the p-value (0.310) is greater than 0.05, so we do not reject the null hypothesis.

Therefore, there is not enough evidence to conclude that the population average IQ is different from 104.