Respuesta :
Answer:
The test statistic is [tex]t = -62.5[/tex]
Step-by-step explanation:
The null hypothesis is:
[tex]H_{0} = 0.2*500 = 100[/tex]
The alternate hypotesis is:
[tex]H_{1} < 0.2*500 < 100[/tex]
Our test statistic is:
[tex]t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
In this problem:
[tex]X = 75, \mu = 100, \sigma = \sqrt{500*0.8*0.2} = 8.94, n = 500[/tex]
So
[tex]t = \frac{75 - 100}{\frac{8.94}{\sqrt{500}}}[/tex]
[tex]t = -62.5[/tex]
The test statistic is [tex]t = -62.5[/tex]
Answer:
[tex]z=\frac{0.15 -0.2}{\sqrt{\frac{0.2(1-0.2)}{500}}}=-2.795[/tex]
Step-by-step explanation:
Data given by the problem
n=500 represent the random sample taken
X=75 represent the students in favor to reducing the deficit using only spending cuts
[tex]\hat p=\frac{75}{500}=0.15[/tex] estimated proportion of favor to reducing the deficit using only spending cuts
[tex]p_o=0.2[/tex] is the value that we want to check
z would represent the statistic
System of hypothesis
We want to cehck if the true proportion of students in favor to reducing the deficit using only spending cuts is less than 0.2.:
Null hypothesis:[tex]p\geq 0.2[/tex]
Alternative hypothesis:[tex]p < 0.2[/tex]
The statistic for this case is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
Replacing the info provided we got:
[tex]z=\frac{0.15 -0.2}{\sqrt{\frac{0.2(1-0.2)}{500}}}=-2.795[/tex]
