Using the z-distribution, as we are working with a proportion, it is found that:
A. Yes, the data provides convincing statistical evidence that Swedish men who would be classified into group A have a lower probability of stroke than Swedish men who would be classified into group B.
B. Considering that the data provides statistical evidence, the conclusion was appropriate.
At the null hypothesis, it is tested if the proportions are the same, that is:
[tex]H_0: p_A - p_B = 0[/tex]
At the alternative hypothesis, it is tested proportion A is less than proportion B, that is:
[tex]H_1: p_A - p_B < 0[/tex]
For each sample, they are given by:
[tex]p_A = \frac{458}{9250} = 0.0495, s_A = \sqrt{\frac{0.0495(0.9505)}{9250}} = 0.0023[/tex]
[tex]p_B = \frac{543}{9250} = 0.0587, s_B = \sqrt{\frac{0.0587(0.9413)}{9250}} = 0.0024[/tex]
Hence, for the distribution of differences, we have that:
[tex]\overline{p} = p_A - p_B = 0.0495 - 0.0587 = -0.0092[/tex]
[tex]s = \sqrt{s_A^2 + s_B^2} = \sqrt{0.0023^2 + 0.0024^2} = 0.0033[/tex]
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{-0.0092 - 0}{0.0033}[/tex]
[tex]z = -2.77[/tex]
Item a:
Considering a left-tailed test, as we are testing if the proportion is less than a value, with the standard significance level of 0.05, the critical value is of [tex]z^{\ast} = -1.645[/tex].
Since the test statistic is less than the critical value, it is found that yes, the data provides convincing statistical evidence that Swedish men who would be classified into group A have a lower probability of stroke than Swedish men who would be classified into group B.
Item b:
Considering that the data provides statistical evidence, the conclusion was appropriate.
More can be learned about the z-distribution at https://brainly.com/question/26454209
