Question

Physicians at a clinic gave what they thought were drugs to 810 patients. Although the doctors later learned that the drugs were really placebos, 58 % of the patients reported an improved condition. Assume that if the placebo is ineffective, the probability of a patients condition improving is 0.53. Test the hypotheses that the proportion of patients improving is >0.53.

Find the test statistics: z =

Find the p-value. p =

Answer 1

The test statistic ( z ) is approximately 3.17, and the p-value is approximately 0.0008.

Step-by-step explanation:

To determine whether the proportion of patients improving differs significantly from 53%, a hypothesis test for a population proportion is conducted. The null hypothesis ( H_0: p \leq 0.53 ) is tested against the alternative hypothesis ( H_1: p > 0.53 ), where ( p ) represents the true proportion of patients whose condition would improve if the placebo were ineffective.

The test statistic ( z ) is calculated using the formula:

`\[ z = \frac{\hat{p} - p_0}{\sqrt{(p_0(1-p_0))/(n)}} \]`

where
`\( \hat{p} \)`is the sample proportion, ( p_0 ) is the hypothesized proportion under the null hypothesis, and ( n ) is the sample size.

Given
`\( \hat{p}` = 0.58 ), ( p_0 = 0.53 ), and the sample size ( n = 810 ), the calculation yields ( z approx 3.17 ).

Next, to find the p-value, we refer to the standard normal distribution. The p-value is the probability of observing a test statistic as extreme as ( z = 3.17 ) under the null hypothesis. Using a standard normal distribution table or statistical software, the p-value is approximately 0.0008. This p-value is less than the common significance levels (such as 0.05 or 0.01), indicating strong evidence against the null hypothesis. Therefore, there is sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis, suggesting that the proportion of patients improving is greater than 53% when given the placebo.