A method currently used by doctors to screen women for possible breast cancer fails to detect cancer in 15% of the women who actually have the disease. A new method has been developed that researchers hope will be able to detect more accurately. A random sample of 80 women who have breast cancer was taken, and the experiment was performed. After the experiment was performed it was discovered that the new method failed to detect the breast cancer in 8 of 80 selected women. Test at 5% level of significance if indeed the new method detects cancer less accurately. Make you decision.

Answer 1

To test if the new method detects breast cancer less accurately than the current method (which fails to detect cancer in 15% of cases), we can set up a hypothesis test using the following steps:

Step 1: Define the Hypotheses

• Null Hypothesis (\(H_0\)): The proportion of failures in the new method is equal to or better than the current method. Mathematically, \(p \geq 0.15\).
• Alternative Hypothesis (\(H_a\)): The proportion of failures in the new method is greater than the current method. Mathematically, \(p > 0.15\).

Step 2: Collect Sample Data

From the experiment, we have:

• Sample size (\(n\)) = 80
• Number of failures in detection (\(x\)) = 8
• Sample proportion of failures (\(\hat{p}\)) = \(\frac{x}{n} = \frac{8}{80} = 0.1\)

Step 3: Determine the Test Statistic

We can use a one-proportion z-test. The test statistic \(z\) is calculated using the formula:

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

Where:

• \(\hat{p}\) = sample proportion of failures
• \(p_0\) = hypothesized population proportion (0.15)
• \(n\) = sample size

Substituting the values:

\[ z = \frac{0.1 - 0.15}{\sqrt{\frac{0.15(1-0.15)}{80}}} \]

Calculating the standard error (SE):

\[ SE = \sqrt{\frac{0.15(0.85)}{80}} = \sqrt{\frac{0.1275}{80}} = \sqrt{0.00159375} \approx 0.0399 \]

Now, substituting back into the z formula:

\[ z = \frac{0.1 - 0.15}{0.0399} = \frac{-0.05}{0.0399} \approx -1.254 \]

Step 4: Find the Critical Value and Make a Decision

Because we are testing at a 5% level of significance for a one-tailed test, we need to find the z critical value for the left tail:

The critical value for \(z\) at the 5% significance level (one-tailed) is approximately -1.645.

Step 5: Compare Test Statistic to Critical Value

• Calculated \(z \approx -1.254\)
• Critical value \(z_{crit} = -1.645\)

Since \(-1.254 > -1.645\), we do not reject the null hypothesis.

Conclusion

The data does not provide sufficient evidence to suggest that the new method detects breast cancer less accurately than the current method at the 5% level of significance. Therefore, we conclude that the new method is not necessarily worse than the current method based on this sample.