A survey claims that 9 out of 10 doctors recommend Aspirin for their patients who have children. To test this claim against an alternative that the actual proportion of doctors who would do this is less than the claimed, a random sample of 100 doctors was considered. In this sample 83 doctors said that they recommend Aspirin for their patients with children. Conduct a hypothesis test using the P-value method.
What conclusions would you make when á= 0.2
To conduct a hypothesis test using the P-value method, we need to set up the null and alternative hypotheses.
Null Hypothesis (H0): The actual proportion of doctors who recommend Aspirin for their patients with children is equal to the claimed proportion (9 out of 10, or 0.9).
Alternative Hypothesis (Ha): The actual proportion of doctors who recommend Aspirin for their patients with children is less than the claimed proportion of 0.9.
Based on the provided data, we have a random sample of 100 doctors, out of which 83 doctors said that they recommend Aspirin for their patients with children.
Next, we need to calculate the test statistic and the corresponding p-value. The test statistic for a proportion can be calculated using the following formula:
z = (p̂ - p0) / sqrt(p0(1-p0) / n)
where p̂ is the sample proportion, p0 is the hypothesized proportion under the null hypothesis, and n is the sample size.
In this case:
p̂ = 83/100 = 0.83 (sample proportion)
p0 = 0.9 (claimed proportion)
n = 100 (sample size)
Plugging these values into the formula, we have:
z = (0.83 - 0.9) / sqrt(0.9*0.1 / 100)
Calculating this, we find that z ≈ -1.36.
To find the p-value, we need to determine the probability of observing a test statistic as extreme as -1.36, assuming the null hypothesis is true. We can do this by consulting a standard normal distribution table or using statistical software.
Using either method, we find that the p-value associated with a test statistic of -1.36 is approximately 0.0869.
Finally, we compare the p-value to the significance level (á) to make our conclusion. In this case, á is given to be 0.2.
Since the p-value (0.0869) is greater than the significance level (0.2), we fail to reject the null hypothesis. This means that we don't have enough evidence to conclude that the actual proportion of doctors who recommend Aspirin for their patients with children is less than the claimed proportion of 0.9.
In conclusion, based on the given data and α = 0.2, we cannot reject the claim that 9 out of 10 doctors recommend Aspirin for their patients with children.