Respuesta :
Answer:
Z = [tex]\frac{0.11-0.08}{\sqrt{0.08*0.92/200} }[/tex] = 1.6
we reject the Null hypothesis
Step-by-step explanation:
The correct test statistic
Assume :
Null Hypothesis : H_0 : p = 0.08
Alternative hypothesis : H_a : p ≠ 0.08
p = proportion of residents that are unemployed
x = unemployed residents = 22
n = sample size = 200
To get the sample of the population that is unemployed we use this equation
[tex]p = \frac{X}{n}[/tex] = 22 / 200 = 0.11
now we apply the test of of significance of single proportion
= Z = [tex]\frac{p - P}{\sqrt{pq/n} }[/tex]
therefore Z = [tex]\frac{0.11-0.08}{\sqrt{0.08*0.92/200} }[/tex] = 1.6
Therefore we reject the Null hypothesis
Testing the hypothesis, it is found that the test statistic used is z = 1.6.
At the null hypothesis, it is tested if the proportion is of 8%, that is:
[tex]H_0: p = 0.08[/tex]
At the alternative hypothesis, it is tested 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}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
In which:
- [tex]\overline{p}[/tex] is the sample proportion.
- p is the proportion tested at the null hypothesis.
- n is the sample size.
For this problem, the parameters are: [tex]p = 0.08, n = 200, \overline{p} = \frac{22}{200} = 0.11[/tex]
Hence, the value of the test statistic is:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
[tex]z = \frac{0.11 - 0.08}{\sqrt{\frac{0.08(0.92)}{200}}}[/tex]
[tex]z = 1.6[/tex]
A similar problem is given at https://brainly.com/question/24166849
