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Midterm Exam 2 Review (Nonadaptive)
Question 5 of 16 (1 point)|Question Attempt: 1 of Unlimited

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Question 5
Partially Correct
(a): Your answer is incorrect.
Use the ALEKS calculator to solve the following problems.
(a)Consider a t distribution with 21 degrees of freedom. Compute P≥t−1.52. Round your answer to at least three decimal places.
P≥t−1.52=
(b)Consider a t distribution with 8 degrees of freedom. Find the value of c such that =P<−c<tc0.99. Round your answer to at least three decimal places.
c=

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To solve this problem, we need to use the properties of the t-distribution for both parts (a) and (b). Here's how to approach each part:

### (a) Consider a t-distribution with 21 degrees of freedom. Compute \( P(t \geq -1.52) \).
We are asked to find the probability \( P(t \geq -1.52) \) for a t-distribution with 21 degrees of freedom. To do this:

1. **Use the cumulative distribution function (CDF) for the t-distribution**:
   - For \( P(t \geq -1.52) \), it is the complement of the CDF at \( t = -1.52 \), i.e.,
     \[
     P(t \geq -1.52) = 1 - P(t \leq -1.52)
     \]
2. Use a calculator or statistical software to find the CDF value for \( t = -1.52 \) with 21 degrees of freedom.

Using a t-distribution table or calculator:
\[
P(t \leq -1.52) \approx 0.072
\]
Thus,
\[
P(t \geq -1.52) = 1 - 0.072 = 0.928
\]

So, the answer to part (a) is:
\[
P(t \geq -1.52) \approx 0.928
\]

### (b) Consider a t-distribution with 8 degrees of freedom. Find the value of \( c \) such that \( P(-c < t < c) = 0.99 \).
This part asks for the critical value \( c \) where 99% of the t-distribution with 8 degrees of freedom lies between \( -c \) and \( c \).

1. **Determine the confidence level**: The given probability is 0.99, so this is a two-tailed probability, and we need to find the value of \( c \) such that:
   \[
   P(-c < t < c) = 0.99
   \]
   This leaves 0.01 split equally in the tails, so each tail has a probability of 0.005.

2. **Find the critical value for the t-distribution**: Using a t-table or a calculator for 8 degrees of freedom and a cumulative probability of \( 0.995 \) (since 0.99 + 0.005 = 0.995 for the upper bound),
   \[
   c \approx 3.355
   \]

Thus, the answer to part (b) is:
\[
c \approx 3.355
\]

### Summary:
- (a) \( P(t \geq -1.52) \approx 0.928 \)
- (b) \( c \approx 3.355 \)

Would you like more details or have any questions?

Here are 5 related questions:
1. How do you compute cumulative probabilities for the t-distribution?
2. What is the difference between one-tailed and two-tailed tests in t-distributions?
3. How does the number of degrees of freedom affect the shape of the t-distribution?
4. Can you explain why we subtract from 1 to find \( P(t \geq x) \)?
5. How do we find critical values using a t-distribution table?

**Tip:** When solving t-distribution problems, always check whether you're dealing with one-tailed or two-tailed probabilities, as this changes the critical value calculation.

Consider a t-distribution with 21 degrees of freedom. Compute P(t ≥ -1.52). Consider a t-distribution with 8 degrees of freedom. Find the value of c such that P(-c < t < c) = 0.99.

Solve t-distribution problems involving degrees of freedom and probability. Compute P(t ≥ -1.52) with 21 degrees of freedom and find the critical value c for P(-c < t < c) = 0.99 with 8 degrees of freedom. Perfect for students in introductory statistics courses.

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