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Early Detection of Breast Cancer Of those women who are diagnosed to have early-stage breast cancer, one-third eventually die of the disease. Suppose a screening program for the early detection of breast cancer was started in order to increase the survival rate \(p\) of those diagnosed to have the disease. A random sample of 200 women was selected from among those who were screened by the program and who were diagnosed to have the disease. Let \(x\) represent the number of those in the sample who survive the disease. a. If you wish to determine whether the screening program has been effective, state the alternative hypothesis that should be tested. b. State the null hypothesis. c. If 164 women in the sample of 200 survive the disease, can you conclude that the screening program was effective? Test using \(\alpha=.05\) and explain the practical conclusions from your test. d. Find the \(p\) -value for the test and interpret it.

Step by step solution

01

Formulate the Hypothesis

We first need to determine the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_1\)). The null hypothesis is that the screening program does not increase the survival rate. The alternative hypothesis is that the program increases the survival rate. $$ H_0: p = \frac{1}{3} \\ H_1: p > \frac{1}{3} $$ where \(p\) is the survival rate of the women who were diagnosed with breast cancer through the screening program.

02

Perform the Hypothesis Test

We will perform a hypothesis test using the \(\alpha=0.05\) significance level. Let's first find the sample proportion of women who survived, which is represented by \(\hat{p}\). $$ \hat{p} = \frac{x}{n} = \frac{164}{200} = 0.82 $$ Next, we will find the test statistic, \(z\), using the following formula: $$ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} $$ where \(p_0 = \frac{1}{3}\), the survival rate in the null hypothesis. $$ z = \frac{0.82 - \frac{1}{3}}{\sqrt{\frac{\frac{1}{3}(1-\frac{1}{3})}{200}}} \approx 21.17 $$ Now, we need to find the critical value for the \(\alpha=0.05\) significance level. Since this is a right-tailed test, we look for the \(z\) critical value, \(z_{\alpha}\), such that the area to the right of it is equal to 0.05. $$ z_{\alpha} = 1.645 $$

03

Compare Test Statistic and Critical Value

Compare the calculated test statistic \(z\) with the critical value \(z_{\alpha}\): - If \(z > z_{\alpha}\), we reject the null hypothesis. - Otherwise, we fail to reject the null hypothesis. In this case, we have: $$ 21.17 > 1.645 $$ Since \(21.17 > 1.645\), we reject the null hypothesis.

04

Practical Conclusion

From the test, we have rejected the null hypothesis and concluded that the survival rate of women who were diagnosed with breast cancer through the screening program is greater than \(\frac{1}{3}\). Therefore, we can say that the screening program is effective in increasing the survival rate.

05

Find the p-value

To find the p-value, we will look up the test statistic \(z\) in a standard normal (z) table. Since our \(z\) value is very large, the p-value will be very small and virtually equal to 0. $$ p\text{-value} \approx 0 $$

06

Interpret the p-value

A p-value of \(\approx 0\) indicates a very low probability that the observed results in the sample could have occurred due to chance alone, assuming the null hypothesis is true. Since the p-value is less than the significance level (\(\alpha = 0.05\)), we can say that the data provides strong evidence against the null hypothesis, supporting the conclusion that the screening program is effective in increasing the survival rate of women diagnosed with breast cancer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Null Hypothesis

The null hypothesis, denoted as \(H_0\), is a statement of no effect or no difference, used as the starting point for hypothesis testing in statistics. In the context of the given exercise, the null hypothesis asserts that the screening program does not improve the survival rates for women diagnosed with early-stage breast cancer. Formally, this is expressed as \(H_0: p = \frac{1}{3}\), where \(p\) represents the true survival rate among these women.

To draw conclusions, one tests the null hypothesis against evidence gathered from sample data. Only if this evidence is sufficiently strong, will the null hypothesis be rejected in favor of the alternative hypothesis. This is a fundamental process in statistical analysis, helpful in determining the effectiveness of treatments or interventions, such as the breast cancer screening program in our example.

Alternative Hypothesis

The alternative hypothesis, denoted as \(H_1\) or \(H_a\), articulates that there is an effect or a difference, and it stands in contrast to the null hypothesis. It's what researchers aim to support with empirical evidence. In our exercise, the alternative hypothesis posits that the screening program increases the survival rate of diagnosed women, symbolically represented by \(H_1: p > \frac{1}{3}\).

If statistical analysis favors the alternative hypothesis, this can lead to breakthroughs in understanding and advancements in medical practices. The alternative hypothesis is crucial to hypothesis testing because it defines the direction and nature of the change a researcher expects to validate.

P-value

The p-value, or probability value, measures the strength of the evidence against the null hypothesis provided by the sample data. It's the probability of observing a test statistic as extreme as, or more extreme than, the one calculated from the sample data, assuming that the null hypothesis is true.

In the breast cancer screening example, the p-value is approximately 0, indicating an extremely low probability that the survival rate observed (82% of the women in the sample survived) is due to random chance if the actual survival rate were indeed as low as one third. A p-value that is lower than the established significance level (\(\alpha\)) suggests that the null hypothesis is unlikely to be true and should be rejected. This concept is pivotal for making data-informed decisions in research and policy-making.

Test Statistic

A test statistic is a standardized value that is calculated from sample data during a hypothesis test. It is used to decide whether to reject the null hypothesis. The test statistic compares the sample data with what is expected under the null hypothesis.

In the exercise, the test statistic is denoted by \(z\), which follows a standard normal distribution and is calculated using the formula given. The resulting value of approximately 21.17 is compared to a critical value that defines the boundary for rejecting the null hypothesis. If the test statistic is greater than the critical value, as it is in this scenario, it provides strong evidence against the null hypothesis. Test statistics are crucial as they translate sample data into an index that can be assessed against a standard to infer conclusions about a population.

Significance Level

The significance level, denoted as \(\alpha\), is the threshold below which the p-value must fall for us to reject the null hypothesis. It provides a basis for decision-making and is chosen by the researcher before data collection begins. Common values for \(\alpha\) are 0.05, 0.01, or 0.10, reflecting a 5%, 1%, or 10% chance of rejecting the null hypothesis when it is true (Type I error).

In our exercise, the significance level was set at 0.05. This means there’s a 5% risk of concluding that the screening program increases survival rates when, in actuality, it does not. Setting an appropriate significance level is important as it reflects the degree of certainty the researcher requires and impacts the conclusions drawn from the hypothesis test.