Respuesta :
Using the z-distribution, it is found that there is enough evidence to conclude that taking a low-dose aspirin each day reduces the chance of developing cancer.
What are the hypotheses tested?
At the null hypotheses, it is tested if there is no difference among the proportions, that is:
[tex]H_0: p_1 - p_2 = 0[/tex]
At the alternative hypotheses, it is tested if the aspirir proportion is lower, that is:
[tex]H_0: p_1 - p_2 < 0[/tex]
What is the mean and the standard error of the distribution of differences?
For each sample, they are given by:
[tex]p_1 = \frac{15}{500} = 0.03, s_1 = \sqrt{\frac{0.03(0.97)}{500}} = 0.0076[/tex]
[tex]p_2 = \frac{26}{500} = 0.052, s_2 = \sqrt{\frac{0.052(0.948)}{500}} = 0.0099[/tex]
Hence, for the distribution of differences, they are given by:
[tex]\overline{p} = p_1 - p_2 = 0.03 - 0.052 = -0.022[/tex]
[tex]s = \sqrt{s_1^2 + s_2^2} = \sqrt{0.0076^2 + 0.0099^2} = 0.0125[/tex]
What is the test statistic?
It is given by:
[tex]z = \frac{\overline{p} - p}{s}[/tex]
In which p = 0 is the value tested at the null hypothesis.
Hence:
[tex]z = \frac{\overline{p} - p}{s}[/tex]
[tex]z = \frac{-0.022 - 0}{0.0125}[/tex]
z = -1.76.
What is the conclusion?
Considering a left-tailed test, as we are testing if the proportion is less than a value, the critical value is [tex]z^{\ast} = -1.645[/tex]
Since the test statistic is less than the critical value for the left-tailed test, there is enough evidence to conclude that taking a low-dose aspirin each day reduces the chance of developing cancer.
More can be learned about the z-distribution at https://brainly.com/question/26454209
