Math Problem Statement

You conduct a test of whether mean home prices in a certain area are $250K or not.   A sample of 12 recent home sales gives an average price of $279K with a standard deviation of $32K. You find that 𝑡=𝑋¯−𝜇0S.E.(𝑋¯)=3.139342t=X¯−μ0S.E.(X¯)=3.139342 You can determine the p-value for you test using which Excel commands below? Select all that work. You might draw pictures and use symmetry to determine the correct selections.

Multiple correct question

=2*T.DIST.RT(3.139342,11)

=2*(1-T.DIST(3.139342,11, TRUE))

=2*T.DIST(-3.139342,11, TRUE)

=2*T.DIST(3.139342,11, TRUE)

Solution

To solve this problem, let's break down the details first:

We are conducting a two-tailed hypothesis test to check whether the mean home price is different from $250K or not. The test statistic tt was calculated as 3.139342 with 11 degrees of freedom (since the sample size is 12, degrees of freedom = 12 - 1 = 11).

Important Considerations:

  • This is a two-tailed test, so we will need to calculate the p-value for both tails of the t-distribution.
  • The Excel commands should correctly calculate the two-tailed p-value, which is 2×P(T>t)2 \times \text{P}(T > t) for a right-tailed value, or equivalently 2×P(T<t)2 \times \text{P}(T < -t) for a left-tailed value.

Let's review the given options:

  1. =2*T.DIST.RT(3.139342,11)

    • T.DIST.RT(t, df) calculates the right-tail probability for the t-distribution.
    • Multiplying by 2 gives the two-tailed p-value, so this is correct for a two-tailed test.
      (Correct)
  2. =2*(1-T.DIST(3.139342,11, TRUE))

    • T.DIST(t, df, TRUE) calculates the cumulative distribution function (CDF), or the area to the left of the given t-value.
    • 1CDF(t)1 - \text{CDF}(t) gives the right-tail probability.
    • Multiplying by 2 makes it a two-tailed test. Hence, this formula is also correct.
      (Correct)
  3. =2*T.DIST(-3.139342,11, TRUE)

    • T.DIST(-t, df, TRUE) calculates the area to the left of a negative t-value.
    • In a symmetric t-distribution, the area to the left of t-t is the same as the area to the right of +t+t.
    • So multiplying this by 2 gives the two-tailed p-value.
      (Correct)
  4. =2*T.DIST(3.139342,11, TRUE)

    • T.DIST(t, df, TRUE) gives the area to the left of the t-value.
    • Multiplying this by 2 would give an incorrect p-value because it includes both the left-tail and the right-tail probability for positive t.
    • This is incorrect for a two-tailed test.
      (Incorrect)

Correct Selections:

  • =2*T.DIST.RT(3.139342,11)
  • =2*(1-T.DIST(3.139342,11, TRUE))
  • =2*T.DIST(-3.139342,11, TRUE)

Would you like more details or have any questions about this explanation?

Here are 5 related questions:

  1. How do you interpret a p-value in hypothesis testing?
  2. What is the relationship between degrees of freedom and the shape of the t-distribution?
  3. What is the difference between one-tailed and two-tailed tests in hypothesis testing?
  4. How do you calculate the test statistic for a t-test?
  5. When should you use a t-test versus a z-test in hypothesis testing?

Tip: When conducting a two-tailed test, always remember to multiply the one-tailed probability by 2 to get the correct p-value.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
t-distribution
p-value Calculation
Two-tailed Test

Formulas

t = (X̄ − μ₀) / S.E.(X̄)
p-value = 2 × P(T > t)
Degrees of Freedom (df) = n - 1

Theorems

t-distribution Symmetry
Central Limit Theorem

Suitable Grade Level

Undergraduate