Questions: Chapter 10Hypothesis Testing: Computations d. Refer to the cancer data referenced in Question 2. Test the hypothesis that the proportion of breast cancer patients treated with ascorbate who will survive more than a year is larger than 0.75. Assume a Type I error rate of a=0.05. Report a p-value, say whether you chose HA or not, and explain why you made the choice you did. Are you surprised by this outcome? Why or why not?

Chapter 10Hypothesis Testing: Computations
d. Refer to the cancer data referenced in Question 2. Test the hypothesis that the proportion of breast cancer patients treated with ascorbate who will survive more than a year is larger than 0.75. Assume a Type I error rate of a=0.05. Report a p-value, say whether you chose HA or not, and explain why you made the choice you did. Are you surprised by this outcome? Why or why not?

Transcript text: Chapter 10|Hypothesis Testing: Computations d. Refer to the cancer data referenced in Question 2. Test the hypothesis that the proportion of breast cancer patients treated with ascorbate who will survive more than a year is larger than 0.75. Assume a Type I error rate of $a=0.05$. Report a $p$-value, say whether you chose $H_{A}$ or not, and explain why you made the choice you did. Are you surprised by this outcome? Why or why not?

Solution

Solution Steps

Step 1: Hypothesis Formulation

We are testing the hypothesis regarding the proportion of breast cancer patients treated with ascorbate who survive more than a year. The null and alternative hypotheses are defined as follows:

Null Hypothesis ($H_0$): $p \leq 0.75$
Alternative Hypothesis ($H_A$): $p > 0.75$

Step 2: Test Statistic Calculation

The test statistic for the proportion is calculated using the formula:

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

Substituting the values:

Sample proportion ($\hat{p}$) = 0.80
Hypothesized proportion ($p_0$) = 0.75
Sample size ($n$) = 100

The calculation yields:

\[ Z = \frac{0.80 - 0.75}{\sqrt{\frac{0.75(1 - 0.75)}{100}}} = 1.1547 \]

Step 3: P-value Calculation

The p-value associated with the calculated test statistic $Z = 1.1547$ is found to be:

\[ \text{P-value} = 0.1241 \]

Step 4: Critical Region and Decision

For a significance level of $\alpha = 0.05$ in a one-tailed test, the critical value is:

\[ Z_{\text{critical}} = 1.6449 \]

Since the calculated test statistic $Z = 1.1547$ does not exceed the critical value, we fail to reject the null hypothesis.

Step 5: Conclusion

Based on the p-value:

Since $0.1241 > 0.05$, we do not have sufficient evidence to support the claim that the proportion of breast cancer patients treated with ascorbate who survive more than a year is larger than 0.75.