Respuesta :
Answer:
[tex]p_v =P(z<-2.298)=0.0107[/tex]
If we compare the pvalue and the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion of surveys returned is not significantly less than 0.2 or 20% .
Step-by-step explanation:
1) Data given and notation
n=6983 represent the random sample taken
X=1322 represent the number of surveys returned
[tex]\hat p=\frac{1322}{6983}=0.189[/tex] estimated proportion of surveys returned
[tex]p_o=0.2[/tex] is the value that we want to test
[tex]\alpha=0.01[/tex] represent the significance level
Confidence=99% or 0.99
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 return rate is less than 20% or 0.2.:
Null hypothesis:[tex]p\geq 0.2[/tex]
Alternative hypothesis:[tex]p <0.2[/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{0.189 -0.2}{\sqrt{\frac{0.2(1-0.2)}{6983}}}=-2.298[/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 provided [tex]\alpha=0.01[/tex]. The next step would be calculate the p value for this test.
Since is a left tailed test the p value would be:
[tex]p_v =P(z<-2.298)=0.0107[/tex]
If we compare the pvalue and the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion of surveys returned is not significantly less than 0.2 or 20% .
