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. Calculate the test statistic.

Answer 1

To test whether the new method detects breast cancer less accurately than the current method, we can formulate our hypothesis and use the results from the sample.

Hypotheses

• Null Hypothesis (\( H_0 \)): The new method detects breast cancer accurately with a failure rate equal to or better than the current method (i.e., the failure rate is \( p \leq 0.15 \)).
• Alternative Hypothesis (\( H_a \): The new method detects breast cancer less accurately than the current method (i.e., the failure rate is \( p > 0.15 \)).

Sample Data

• Total women sampled (\( n \)) = 80
• Number of women in which the new method failed to detect breast cancer = 8

The sample proportion (\( \hat{p} \)) of the new method's failure rate can be calculated as: \[ \hat{p} = \frac{\text{Number of failures}}{\text{Total sample size}} = \frac{8}{80} = 0.1 \]

Test Statistic

We can use the Z-test for proportions. The test statistic \( Z \) is calculated as follows:

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

where:

• \( \hat{p} \) is the sample proportion (0.1),
• \( p_0 \) is the hypothesized population proportion (0.15),
• \( n \) is the sample size (80).

Now, we need to calculate the standard error using \( p_0 \):

\[ \text{Standard Error} = \sqrt{\frac{p_0(1 - p_0)}{n}} = \sqrt{\frac{0.15(1 - 0.15)}{80}} = \sqrt{\frac{0.15 \times 0.85}{80}} = \sqrt{\frac{0.1275}{80}} = \sqrt{0.00159375} \approx 0.0399 \]

Calculating the Z-Statistic

Now we substitute the values:

\[ Z = \frac{0.1 - 0.15}{0.0399} = \frac{-0.05}{0.0399} \approx -1.253 \]

Conclusion

The calculated Z-statistic is approximately \(-1.253\).

Next, we will compare this value to the critical Z-value for a one-tailed test at the 5% level of significance. The critical Z-value (which you can find in the Z-table) for a one-tailed test at \( \alpha = 0.05 \) is approximately \( 1.645 \).

Since our calculated Z-value of \(-1.253\) is greater than \(-1.645\), we do not reject the null hypothesis. This means that there is not enough evidence at the 5% level of significance to conclude that the new method detects cancer less accurately than the current method.

Final Answer: The test statistic is approximately -1.253.