(b)Consider a *t* distribution with 26 degrees of freedom. Find the value of c such that P(t less than or equal to c)=0.05. Round your answer to at least three decimal places.

To find the value of \( c \) such that \( P(t \leq c) = 0.05 \) for a t-distribution with 26 degrees of freedom, you need to find the 5th percentile of this distribution. This is typically done using the inverse cumulative distribution function (inverse CDF), also known as the percent point function (PPF).

I'll calculate this value now.The value of \( c \) such that \( P(t \leq c) = 0.05 \) for a t-distribution with 26 degrees of freedom is approximately **-1.706**.

Would you like to explore this further or ask any other questions?

Here are some related questions you might be interested in:
1. How do you interpret the 5th percentile in a t-distribution?
2. What is the difference between a one-tailed and two-tailed t-test?
3. How does the t-distribution compare to the normal distribution for larger degrees of freedom?
4. How can the inverse CDF (PPF) be used in hypothesis testing?
5. What are critical values in the context of t-tests, and how are they used?
6. How does changing the degrees of freedom affect the critical value in a t-distribution?
7. How can you find the critical value for a two-tailed test at the 95% confidence level?
8. What is the relationship between p-values and critical values in hypothesis testing?

**Tip:** When dealing with t-distributions, knowing how to find critical values is crucial for conducting hypothesis tests and constructing confidence intervals.

Learn how to find the value of c for a t-distribution with 26 degrees of freedom, such that P(t ≤ c) = 0.05. This involves calculating the 5th percentile using the inverse cumulative distribution function (inverse CDF). Suitable for advanced high school and undergraduate students studying statistics and probability.

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