Respuesta :
Answer:
[tex]z=\frac{0.48 -0.47}{\sqrt{\frac{0.47(1-0.47)}{1004}}}=0.635[/tex]
[tex]p_v =P(z>0.635)=1-P(z<0.635)=1-0.737=0.263[/tex]
Step-by-step explanation:
1) Data given and notation n
n=1004 represent the random sample taken
X=482 represent the Chevrolet owners who said they would buy another Chevrolet.
[tex]\hat p=\frac{482}{1004}=0.480[/tex] estimated proportion of Chevrolet owners who said they would buy another Chevrolet.
[tex]p_o=0.47[/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 the proportion of Chevrolet owners who said they would buy another Chevrolet is higher than 47%:
Null hypothesis:[tex]p\leq 0.47[/tex]
Alternative hypothesis:[tex]p > 0.47[/tex]
When we conduct a proportion test we need to use the z statisitc, 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{0.48 -0.47}{\sqrt{\frac{0.47(1-0.47)}{1004}}}=0.635[/tex]
4) Statistical decision
P value method or p value approach . "This method consists on 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.
If we use the significance level provided, [tex]\alpha=0.01[/tex]. The next step would be calculate the p value for this test.
Since is a one side upper test the p value would be:
[tex]p_v =P(z>0.635)=1-P(z<0.635)=1-0.737=0.263[/tex]
So based on the p value obtained and using the significance level assumed [tex]\alpha=0.01[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion of Chevrolet owners who said they would buy another Chevroletis not significantly higher than 0.47 or 47% .
