Respuesta :
Answer:
[tex]z=\frac{0.710 -0.68}{\sqrt{\frac{0.68(1-0.68)}{995}}}=2.029[/tex]
The p value would be given by:
[tex]p_v =P(z>2.029)=0.021[/tex]
In the ti84 we follow these steps:
STAT, TESTS, 1Pro-Test
And we put the following input:
po= 0.68
x= 706, n=995 and prop [tex]>p_o[/tex]
And then click on Calculate
If the significance level is higher than the p value we can reject the null hypothesis otherwise no.
Step-by-step explanation:
Information given
n=995 represent the random sample taken
X=706 represent the number of people who support federal legislation putting limits on the amounts that top executives are paid at companies that receive emergency government loans
[tex]\hat p=\frac{706}{995}=0.710[/tex] estimated proportion of interest
[tex]p_o=0.68[/tex] is the value that we want to test
z would represent the statistic
[tex]p_v[/tex] represent the p value
Hypothesis to test
We want to test if the true proportion is higher than 0.68.:
Null hypothesis:[tex]p\leq 0.68[/tex]
Alternative hypothesis:[tex]p > 0.68[/tex]
The statistic is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
Replacing the info given we got:
[tex]z=\frac{0.710 -0.68}{\sqrt{\frac{0.68(1-0.68)}{995}}}=2.029[/tex]
The p value would be given by:
[tex]p_v =P(z>2.029)=0.021[/tex]
In the ti84 we follow these steps:
STAT, TESTS, 1Pro-Test
And we put the following input:
po= 0.68
x= 706, n=995 and prop [tex]>p_o[/tex]
And then click on Calculate
If the significance level is higher than the p value we can reject the null hypothesis otherwise no.
