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)

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 $ t $ 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 \times \text{P}(T > t) $ for a right-tailed value, or equivalently $ 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.
   - $ 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 $-t$ is the same as the area to the right of $+t$.
   - 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.

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 = X̄ − μ₀ / S.E.(X̄) = 3.139342. You can determine the p-value for your test using which Excel commands below? Select all that work.

=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)

Learn how to test whether mean home prices in a specific area differ from $250K using hypothesis testing with t-distribution. This problem involves calculating the p-value from a sample of 12 home sales with an average price of $279K and standard deviation of $32K. Suitable for undergraduate students.

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