Respuesta :
Answer:
a) [tex]p_v =P(Z>1.119)=0.132[/tex]
So the p value obtained was a high low value and using the significance level assumed [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.
b) Since we Fail to reject the null hypothesis and the 0.5 is included on the null hypothesis we would expect that 0.5 would be on the confidence interval.
The confidence interval would be given by this formula
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For the 95% confidence interval the value of [tex]\alpha=1-0.5=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.96[/tex]
And replacing into the confidence interval formula we got:
[tex]0.52 - 1.96 \sqrt{\frac{0.52(1-0.52)}{783}}=0.485[/tex]
[tex]0.52 + 1.96 \sqrt{\frac{0.52(1-0.52)}{783}}=0.555[/tex]
Step-by-step explanation:
1) Data given and notation
n=783 represent the random sample taken
X represent the people with the characteristic of interest
[tex]\hat p=0.52[/tex] estimated proportion of independents who oppose the plan
[tex]p_o=0.5[/tex] is the value that we want to test
[tex]\alpha[/tex] represent the significance level
z would represent the statistic (variable of interest)
[tex]p_v{/tex} represent the p value (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that that a majority of Independents oppose the health care public option plan:
Null hypothesis:[tex]p \leq 0.5[/tex]
Alternative hypothesis:[tex]p > 0.5[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.52 -0.5}{\sqrt{\frac{0.5(1-0.5)}{783}}}=1.119[/tex]
4) Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level assumed is [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a right tailed test the p value would be:
[tex]p_v =P(Z>1.119)=0.132[/tex]
So the p value obtained was a high low value and using the significance level assumed [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.
Part b
Since we Fail to reject the null hypothesis and the 0.5 is included on the null hypothesis we would expect that 0.5 would be on the confidence interval.
The confidence interval would be given by this formula
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For the 95% confidence interval the value of [tex]\alpha=1-0.5=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.96[/tex]
And replacing into the confidence interval formula we got:
[tex]0.52 - 1.96 \sqrt{\frac{0.52(1-0.52)}{783}}=0.485[/tex]
[tex]0.52 + 1.96 \sqrt{\frac{0.52(1-0.52)}{783}}=0.555[/tex]