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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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Solution

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(t1.52)P(t \geq -1.52).

We are asked to find the probability P(t1.52)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(t1.52)P(t \geq -1.52), it is the complement of the CDF at t=1.52t = -1.52, i.e., P(t1.52)=1P(t1.52)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=1.52t = -1.52 with 21 degrees of freedom.

Using a t-distribution table or calculator: P(t1.52)0.072P(t \leq -1.52) \approx 0.072 Thus, P(t1.52)=10.072=0.928P(t \geq -1.52) = 1 - 0.072 = 0.928

So, the answer to part (a) is: P(t1.52)0.928P(t \geq -1.52) \approx 0.928

(b) Consider a t-distribution with 8 degrees of freedom. Find the value of cc such that P(c<t<c)=0.99P(-c < t < c) = 0.99.

This part asks for the critical value cc where 99% of the t-distribution with 8 degrees of freedom lies between c-c and cc.

  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 cc such that: P(c<t<c)=0.99P(-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 0.9950.995 (since 0.99 + 0.005 = 0.995 for the upper bound), c3.355c \approx 3.355

Thus, the answer to part (b) is: c3.355c \approx 3.355

Summary:

  • (a) P(t1.52)0.928P(t \geq -1.52) \approx 0.928
  • (b) c3.355c \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(tx)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.

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Math Problem Analysis

Mathematical Concepts

Statistics
T-distribution
Probability

Formulas

P(t ≥ -1.52) = 1 - P(t ≤ -1.52)
P(-c < t < c) = 0.99

Theorems

T-distribution cumulative probability theorem
Two-tailed test for confidence intervals

Suitable Grade Level

Undergraduate Level (Introductory Statistics)