Consider a permutation test of the null hypothesis that there is no effect of the mutagen, i.e. that the mean number of microsatellite nuclei is the same in the two groups. We will use the difference in means (dose of 80 mg/ml - control) as our test statistic. So, a positive value for the test statistic indicates that the 80 mg/ml dose had a larger number of microsatellite nuclei. The observed difference (calculated from the data above) is 8.2
There are a total of 126 permutations of 9 things into a group of 4 and group of 5. The 126 values of the test statistic (difference in means as dose - no), sorted from smallest to largest, are:
-9.35 -8.9 -8 -8 -8 -8 -8 -7.1 -7.1 -7.1 -7.1 -6.65 -6.65 -6.65 -6.65 -5.75 -5.75 -5.75 -5.75 -5.75 -5.75 -5.3 -4.4 -4.4 -4.4 -4.4 -3.95 -3.95 -3.95 -3.95 -3.05 -3.05 -3.05 -3.05 -3.05 -3.05 -3.05 -3.05 -3.05 -3.05 -2.15 -2.15 -2.15 -2.15 -2.15 -2.15 -1.7 -1.7 -1.7 -1.7 -1.7 -1.7 -1.25 -0.8 -0.8 -0.8 -0.8 -0.35 -0.35 -0.35 -0.35 0.1 0.1 0.1 0.1 1 1 1 1 1 1 1 1 1 1 1 1.9 1.9 1.9 1.9 1.9 1.9 2.35 2.35 2.35 2.35 2.35 2.35 3.25 3.25 3.25 3.25 3.7 3.7 3.7 3.7 4.6 4.6 4.6 4.6 4.6 4.6 5.05 5.05 5.05 5.05 5.05 5.05 5.95 5.95 5.95 5.95 5.95 5.95 5.95 5.95 5.95 5.95 6.85 6.85 6.85 6.85 7.3 7.3 7.3 8.2
a) If the null hypothesis (no difference) is true, what is the probability of 8.2 or a more extreme positive value?
b) What is the two-sided p-value for the test of the null hypothesis of no difference?

Let's break this down step by step to address each question.

### (a) Probability of \( 8.2 \) or a more extreme positive value

In permutation tests, the **p-value** is calculated as the proportion of the permutation test statistics that are as extreme or more extreme than the observed test statistic. The observed test statistic in this case is \( 8.2 \).

To calculate the probability of getting \( 8.2 \) or a more extreme positive value, we need to count how many of the 126 permutations have a test statistic of \( 8.2 \) or greater. 

Looking at the sorted list of test statistics:
- The values equal to or greater than \( 8.2 \) are only \( 8.2 \) itself.

Since there is only one permutation with a value of \( 8.2 \), the probability is:

\[
\text{Probability} = \frac{\text{Number of values} \geq 8.2}{\text{Total number of permutations}} = \frac{1}{126}
\]

So, the probability of observing \( 8.2 \) or a more extreme positive value under the null hypothesis is \( \frac{1}{126} \approx 0.00794 \).

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### (b) Two-sided p-value

A **two-sided p-value** considers both the extreme positive and extreme negative values. To calculate this, we need to find how many of the 126 permutations are as extreme as or more extreme than \( 8.2 \) (on the positive side) and its negative counterpart \( -8.2 \) (on the negative side).

From the sorted list, we can count:
- There is one value equal to \( 8.2 \).
- On the negative side, the smallest value is \( -9.35 \), and the negative values more extreme than \( -8.2 \) are \( -9.35, -8.9, -8, -8, -8, -8, -8 \). There are **7 values** that are \( \leq -8 \).

Thus, there are \( 1 + 7 = 8 \) values that are as extreme as or more extreme than \( 8.2 \) or \( -8.2 \).

The two-sided p-value is:

\[
\text{Two-sided p-value} = \frac{\text{Number of values } \leq -8.2 \text{ or } \geq 8.2}{\text{Total number of permutations}} = \frac{8}{126} \approx 0.0635
\]

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### Summary:
- **(a)** The probability of observing \( 8.2 \) or a more extreme positive value is \( \frac{1}{126} \approx 0.00794 \).
- **(b)** The two-sided p-value is \( \frac{8}{126} \approx 0.0635 \).

Would you like further details or clarifications on any steps?

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Here are 5 related questions you might find useful:
1. How is a permutation test different from a t-test?
2. What are the advantages of using a permutation test?
3. How would the results change if we used a different test statistic (e.g., median difference)?
4. How does the size of the sample affect the results of a permutation test?
5. Can permutation tests be used for more than two groups?

**Tip**: Permutation tests are a powerful non-parametric method because they don’t rely on normal distribution assumptions!

Consider a permutation test of the null hypothesis that there is no effect of the mutagen, i.e. that the mean number of microsatellite nuclei is the same in the two groups. We will use the difference in means (dose of 80 mg/ml - control) as our test statistic. So, a positive value for the test statistic indicates that the 80 mg/ml dose had a larger number of microsatellite nuclei. The observed difference (calculated from the data above) is 8.2. There are a total of 126 permutations of 9 things into a group of 4 and group of 5. The 126 values of the test statistic (difference in means as dose - no), sorted from smallest to largest, are provided. (a) If the null hypothesis (no difference) is true, what is the probability of 8.2 or a more extreme positive value? (b) What is the two-sided p-value for the test of the null hypothesis of no difference?

This problem explores how to conduct a permutation test for the null hypothesis of no effect of a mutagen on microsatellite nuclei count. We calculate the probability of observing a difference in means of 8.2 or more extreme, along with the two-sided p-value. Step-by-step calculations and insights into non-parametric hypothesis testing.

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