Respuesta :
Answer:
(a) The z-statistic for this test is 1.265.
(b) The P-Value is 0.102936.
Step-by-step explanation:
Given information:
n=1000
Republican candidate = 520
Sample proportion p = [tex]\frac{520}{1000}=0.52[/tex]
H0 : p=0.50
Ha : p>0.50
It is right tailed test.
Hypothesized proportion = 0.5
Assumed that the data is normally distributed.
(a) The z-statistic for this test is
[tex]z=\frac{p-P}{\sqrt{\frac{PQ}{n}}}[/tex]
where, p is sample proportion .
P is hypothesized proportion.
Q = 1-P
n is sample size.
The z-statistic for this test is
[tex]z=\frac{0.52-0.5}{\sqrt{\frac{0.5(1-0.5}{1000}}}[/tex]
[tex]z=1.26491106407[/tex]
[tex]z\approx 1.265[/tex]
Therefore the z-statistic for this test is 1.265.
(b)
Level of significance is not given so let as consider α =0.05.
The P-Value for right tailed test at significance level α =0.05 is 0.102936.
Therefore P-Value is 0.102936.
Using the z-distribution, as we are working with a proportion, it is found that:
a) The z-statistic for this test is of z = 1.265.
b) The p-value for this test is of 0.1029.
What is the z-statistic?
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.
Item a:
In this problem, the parameters are:
[tex]n = 1000, p = 0.5, \overline{p} = \frac{520}{1000} = 0.52[/tex].
Hence:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
[tex]z = \frac{0.52 - 0.5}{\sqrt{\frac{0.5(0.5)}{1000}}}[/tex]
[tex]z = 1.265[/tex]
The z-statistic for this test is of z = 1.265.
Item b:
Using a z-distribution calculator, with z = 1.265 and a right-tailed test, as we are testing if the proportion is greater than a value, it is found that:
The p-value for this test is of 0.1029.
More can be learned about the z-distribution at https://brainly.com/question/26454209
