Question

Directions: Show all your work. Indicate clearly the methods you use, because you will be scored on the correctness of your methods as well as on the accuracy and completeness of your results and explanations. A researcher conducted a medical study to investigate whether taking a low-dose aspirin reduces the chance of developing colon cancer. As part of the study, 1,000 adult volunteers were randomly assigned to one of two groups. Half of the volunteers were assigned to the experimental group that took a low-dose aspirin each day, and the other half were assigned to the control group that took a placebo each day. At the end of six years, 15 of the people who took the low-dose aspirin had developed colon cancer and 26 of the people who took the placebo had developed colon cancer. At the significance level α = 0.05, do the data provide convincing statistical evidence that taking a low-dose aspirin each day would reduce the chance of developing colon cancer among all people similar to the volunteers

Answer 1

Answer:

Step-by-step explanation:

To determine if taking a low-dose aspirin reduces the chance of developing colon cancer, we will conduct a hypothesis test using the provided data.

Let's define our hypotheses:

Null Hypothesis (H0): Taking a low-dose aspirin does not reduce the chance of developing colon cancer.

Alternative Hypothesis (H1): Taking a low-dose aspirin reduces the chance of developing colon cancer.

We will use a two-proportion z-test to compare the proportions of colon cancer development between the two groups (aspirin and placebo).

First, let's calculate the proportions of colon cancer development for each group:

Experimental group (aspirin): 15 out of 500 volunteers developed colon cancer.

Control group (placebo): 26 out of 500 volunteers developed colon cancer.

Now, we will calculate the test statistic (z-score) using the formula:

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

Where:

\( p_1 \) = proportion of colon cancer development in the experimental group

\( p_2 \) = proportion of colon cancer development in the control group

\( p \) = combined proportion of colon cancer development

\( n_1 \) = sample size of the experimental group

\( n_2 \) = sample size of the control group

Let's calculate \( p_1 \), \( p_2 \), and \( p \):

\( p_1 = \frac{15}{500} = 0.03 \) (proportion of colon cancer development in the experimental group)

\( p_2 = \frac{26}{500} = 0.052 \) (proportion of colon cancer development in the control group)

\( p = \frac{15 + 26}{1000} = \frac{41}{1000} = 0.041 \) (combined proportion of colon cancer development)

Now, we can calculate the z-score:

\[ z = \frac{(0.03 - 0.052)}{\sqrt{0.041(1-0.041)(\frac{1}{500} + \frac{1}{500})}} \]

\[ z = \frac{(-0.022)}{\sqrt{0.041(0.959)(0.004)}} \]

\[ z \approx \frac{(-0.022)}{\sqrt{0.01583}} \]

\[ z \approx \frac{(-0.022)}{0.1259} \]

\[ z \approx -0.1748 \]

Next, we will find the critical value from the standard normal distribution table for a significance level of \( \alpha = 0.05 \). Since this is a two-tailed test, we divide \( \alpha \) by 2 to get \( \alpha/2 = 0.025 \).

Using the standard normal distribution table, the critical value for \( \alpha/2 = 0.025 \) is approximately -1.96.

Since the absolute value of the calculated z-score (0.1748) is less than the critical value (-1.96), we fail to reject the null hypothesis.

Conclusion: At the significance level \( \alpha = 0.05 \), there is not enough evidence to conclude that taking a low-dose aspirin each day reduces the chance of developing colon cancer among all people similar to the volunteers.