Let's solve each part of this problem:

Part (a)

We are given a $t$-distribution with 3 degrees of freedom, and we need to calculate: $P(t \leq -1.89)$ To find this probability, we use the cumulative distribution function (CDF) for the $t$-distribution with 3 degrees of freedom at $t = -1.89$. Typically, this would involve looking up this value in a $t$-distribution table or using statistical software.

Using a calculator or software: $P(t \leq -1.89) \approx 0.086$ (rounded to three decimal places).

Part (b)

For this part, we have a $t$-distribution with 13 degrees of freedom, and we need to find the value of $c$ such that: $P(-c < t < c) = 0.95$ This problem is asking for the critical value $c$ that captures the middle 95% of the distribution. For a $t$-distribution with 13 degrees of freedom, we look up the value of $c$ for the 0.975 quantile (since 0.975 on one side and 0.025 on the other add to 0.95).

Using a calculator or statistical software: $c \approx 2.160$ (rounded to three decimal places).

Summary of Results

• (a) $P(t \leq -1.89) \approx 0.086$
• (b) $c \approx 2.160$

Would you like more detailed explanations, or do you have any questions?

Related Questions

1. How does the degrees of freedom affect the shape of the $t$-distribution?
2. What is the difference between a $t$-distribution and a standard normal distribution?
3. How would you compute $P(t \leq x)$ for other values of $x$ in a $t$-distribution?
4. Why is the $t$-distribution used instead of the normal distribution when the sample size is small?
5. What are critical values and how are they used in hypothesis testing?

Tip

For critical value calculations in a $t$-distribution, remember that the confidence level and degrees of freedom determine the critical $t$-values used in various statistical tests.