Respuesta :
Answer:
Null hypothesis:[tex]p=0.114[/tex]
Alternative hypothesis:[tex]p \neq 0.114[/tex]
[tex]z=\frac{0.152 -0.114}{\sqrt{\frac{0.114(1-0.114)}{850}}}=3.49[/tex]
Since is a bilateral test the p value would be:
[tex]p_v =2*P(z>3.49)=0.00049[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis.
[tex]0.152 - 1.96 \sqrt{\frac{0.152(1-0.152)}{850}}=0.128[/tex]
[tex]0.152 - 1.96 \sqrt{\frac{0.152(1-0.152)}{850}}=0.176[/tex]
And the 95% confidence interval would be given (0.128;0.176).
And support the conclusion obtained on the hypothesis test since the value of 0.114 is not in the confidence interval, so we have enough evidence to reject the null hypothesis.
Step-by-step explanation:
Data given and notation
n=850 represent the random sample taken
X=129 represent the number of housing units that are vacant.
[tex]\hat p=\frac{129}{850}=0.152[/tex] estimated proportion of housing units that are vacant.
[tex]p_o=0.114[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
Confidence=95% or 0.95
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the proportion is equal is 0.114 or not:
Null hypothesis:[tex]p=0.114[/tex]
Alternative hypothesis:[tex]p \neq 0.114[/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].
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.152 -0.114}{\sqrt{\frac{0.114(1-0.114)}{850}}}=3.49[/tex]
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 bilateral test the p value would be:
[tex]p_v =2*P(z>3.49)=0.00049[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis.
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=1-0.95=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]0.152 - 1.96 \sqrt{\frac{0.152(1-0.152)}{850}}=0.128[/tex]
[tex]0.152 - 1.96 \sqrt{\frac{0.152(1-0.152)}{850}}=0.176[/tex]
And the 95% confidence interval would be given (0.128;0.176).
And support the conclusion obtained on the hypothesis test since the value of 0.114 is not in the confidence interval, so we have enough evidence to reject the null hypothesis.
