Respuesta :
Given the question:
The appropriate hypotheses for the question: do these data provide evidence that the 8% value is inaccurate is given by
[tex]H_o:p=0.08 \\ \\ H_a:p\neq0.08[/tex]
Part B:
The proportion of children in this sample that are nearsighted is given by
[tex] \frac{21}{194} =0.1082=10.82\%[/tex]
Part C
Given that the standard error of the sample proportion is 0.0195 and the point estimate follows a nearly normal distribution,
The test statistic is calculated as follows:
[tex] \frac{\hat{p}-p}{SE_p} = \frac{0.1082-0.08}{0.0195} = \frac{0.0282}{0.0195} =1.446[/tex]
Therefore, the test statistic is 1.446.
Part D
The p-value for this hypothesis test is given by
[tex]P(-1.446 \leq z \leq 1.446)=2P(z\leq-1.446)=2(0.0741)=0.1482[/tex]
Part E
Since the P-value (0.1482) is relatively large, we cannot reject the null hypothesis.
Therefore, we conclude that these data does not provide evidence that the 8% value is inaccurate.
Nearsighted. It is believed that nearsightedness affects about 8% of all children. In a random sample of 194 children, 21 are nearsighted.
(a) Construct hypotheses appropriate for the following question: do these data provide evidence that the 8% value is inaccurate?
(b) What proportion of children in this sample are nearsighted?
(c) Given that the standard error of the sample proportion is 0.0195 and the point estimate follows a nearly normal distribution, calculate the test statistic (the Z-statistic).
(d) What is the p-value for this hypothesis test?
(e) What is the conclusion of the hypothesis test?
Part A:The appropriate hypotheses for the question: do these data provide evidence that the 8% value is inaccurate is given by
[tex]H_o:p=0.08 \\ \\ H_a:p\neq0.08[/tex]
Part B:
The proportion of children in this sample that are nearsighted is given by
[tex] \frac{21}{194} =0.1082=10.82\%[/tex]
Part C
Given that the standard error of the sample proportion is 0.0195 and the point estimate follows a nearly normal distribution,
The test statistic is calculated as follows:
[tex] \frac{\hat{p}-p}{SE_p} = \frac{0.1082-0.08}{0.0195} = \frac{0.0282}{0.0195} =1.446[/tex]
Therefore, the test statistic is 1.446.
Part D
The p-value for this hypothesis test is given by
[tex]P(-1.446 \leq z \leq 1.446)=2P(z\leq-1.446)=2(0.0741)=0.1482[/tex]
Part E
Since the P-value (0.1482) is relatively large, we cannot reject the null hypothesis.
Therefore, we conclude that these data does not provide evidence that the 8% value is inaccurate.