To find the p-value for this sample, we need to conduct a one-sample t-test since the population standard deviation is unknown and the sample size is small ($n = 18$).

Step 1: State the hypotheses

• Null hypothesis ($H_0$): $\mu = 89.9$
• Alternative hypothesis ($H_a$): $\mu < 89.9$ (This is a one-tailed test)

Step 2: Calculate the t-statistic

The t-statistic is calculated using the following formula:

$t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}$

Where:

• $\bar{x} = 83.9$ (sample mean)
• $\mu_0 = 89.9$ (hypothesized population mean)
• $s = 14.1$ (sample standard deviation)
• $n = 18$ (sample size)

Substitute the values into the formula:

$t = \frac{83.9 - 89.9}{\frac{14.1}{\sqrt{18}}}$

$t = \frac{-6}{\frac{14.1}{\sqrt{18}}}$

$t \approx \frac{-6}{3.323} \approx -1.805$

Step 3: Determine the degrees of freedom

The degrees of freedom ($df$) for this t-test is:

$df = n - 1 = 18 - 1 = 17$

Step 4: Find the p-value

Using the t-distribution table or a calculator, we find the p-value associated with the t-statistic $-1.805$ with 17 degrees of freedom.

$\text{p-value} \approx 0.0441$

Conclusion

The p-value for this sample is approximately 0.0441 (to four decimal places).

Do you want details or have any questions?

---

1. What is the significance level and how does it relate to the p-value?
2. How does the sample size affect the t-statistic and p-value?
3. What are degrees of freedom and how are they calculated?
4. What does a one-tailed test mean in hypothesis testing?
5. How can you use the t-distribution table to find the p-value?

Tip: When the population standard deviation is unknown and the sample size is small, the t-distribution should be used instead of the normal distribution.