Respuesta :
Answer:
These data do not provide evidence that the 8% value is inaccurate, this at the significance level of 5%.
Step-by-step explanation:
Let p be the true proportion of children nearsighted. We want to test the next hypothesis:
[tex]H_{0}: p = 0.08[/tex] vs [tex]H_{a}: p \neq 0.08[/tex] (two-tailed alternative)
We have a large sample size of n = 194 children. Therefore, the test statistic is given by
Â
[tex]Z = \frac{\hat{p}-0.08}{\sqrt{0.08(0.92)/194}}[/tex] which is normally distributed.
The observed value is Â
[tex]z = \frac{21/194-0.08}{\sqrt{0.08(0.92)/194}} = 1.4502[/tex]. Â
The rejection region using a significance level of 0.05 is given by RR = {z | z < -1.96 or z > 1.96}. Because the observed value 1.4502 does not fall inside the rejection region, we fail to reject the null hypothesis.
Testing the hypothesis, we can conclude that since the p-value of the test is 0.147, which is greater than the standard significance level of 0.05, thus, these data does not provide evidence that the 8% value is inaccurate.
At the null hypothesis, we test if the proportion is of 8%, that is:
[tex]H_0: p = 0.08[/tex]
At the alternative hypothesis, we test if the proportion is different of 8%, that is:
[tex]H_1: p \neq 0.08[/tex]
The test statistic is given by:
[tex]z = \frac{\overline{p} - p}{s}[/tex]
In which:
[tex]s = \sqrt{\frac{p(1 - p)}{n}}[/tex]
In this problem:
- 8% is tested at the null hypothesis, thus [tex]p = 0.08[/tex].
- 21 out of 194, thus [tex]n = 194, \overline{p} = \frac{21}{194} = 0.10825[/tex].
Then, the value of the test statistic is:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1 - p)}{n}}}[/tex]
[tex]z = \frac{0.10825 - 0.08}{\sqrt{\frac{0.08(0.92)}{194}}}[/tex]
[tex]z = 1.45[/tex]
The p-value of the test, since we are doing a two-tailed test(testing if the mean is different from a value), is P(|z| > 1.45), which is 2 multiplied by the p-value of z = -1.45.
Looking at the z-table, z = -1.45 has a p-value of 0.0735.
2(0.0735) = 0.147
The p-value of the test is 0.147, which is greater than the standard significance level of 0.05, thus, these data does not provide evidence that the 8% value is inaccurate.
A similar problem is given at https://brainly.com/question/24330815
