Question

There was a total of 130 participants in the placebo group, of which 70% reported a major decrease in their IBS severity score. The no-pill control group had 87 participants, of whom 54% reported a major decrease in their IBS severity score.

Is this evidence that taking placebo pills is more likely to result in a major decrease in the IBS severity score than taking no pills? Set up a test of significance to answer this question.

Answer 1

After calculating z, we can compare it to the critical value from the standard normal distribution or find the p-value. If the result is statistically significant at the chosen significance level (e.g., 0.05), we would reject the null hypothesis and conclude that there is evidence that taking placebo pills is more likely to result in a major decrease in the IBS severity score than taking no pills.

To determine if taking placebo pills is more likely to result in a major decrease in the IBS severity score than taking no pills, we can perform a hypothesis test for the difference in proportions.

Let:

-
`\( p_1 \)` be the proportion of the placebo group reporting a major decrease.

-
`\( p_2 \)` be the proportion of the no-pill control group reporting a major decrease.

The null hypothesis
`(\(H_0\))` is that there is no difference between the two proportions, and the alternative hypothesis
`(\(H_a\))` is that the proportion of those reporting a major decrease is higher in the placebo group.

The hypotheses are:

`\[ H_0: p_1 = p_2 \]`

`\[ H_a: p_1 > p_2 \]`

Now, let's perform a z-test for the difference in proportions.

`\[ z = \frac{(p_1 - p_2)}{\sqrt{p(1-p)((1)/(n_1) + (1)/(n_2))}} \]`

Where:

- p is the pooled sample proportion.

-
`\( n_1 \)` is the sample size of the placebo group.

-
`\( n_2 \)` is the sample size of the no-pill control group.

The pooled sample proportion is calculated as:

`\[ p = (X_1 + X_2)/(n_1 + n_2) \]`

where
`\( X_1 \)` and
`\( X_2 \)` are the total number of successes in each group.

In this case:

`\[ X_1 = 130 * 0.70 = 91 \]` (number of successes in the placebo group)

`\[ X_2 = 87 * 0.54 = 47 \]` (number of successes in the no-pill control group)

`\[ p = (91 + 47)/(130 + 87) \approx 0.613 \]`

Now, plug in the values to calculate the z-statistic and compare it to the critical value or find the p-value.

`\[ z = \frac{(0.70 - 0.54)}{\sqrt{0.613(1-0.613)((1)/(130) + (1)/(87))}} \]`

After calculating z, we can compare it to the critical value from the standard normal distribution or find the p-value. If the result is statistically significant at the chosen significance level (e.g., 0.05), we would reject the null hypothesis and conclude that there is evidence that taking placebo pills is more likely to result in a major decrease in the IBS severity score than taking no pills.