Respuesta :
Answer:
[tex]z=\frac{0.38 -0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}=-2.4[/tex]
[tex]p_v =P(z<-2.4)=0.0082[/tex]
And we can use the following excel code to find it:
"=NORM.DIST(-2.4,0,1,TRUE)"
Step-by-step explanation:
Data given and notation
n=100 represent the random sample taken
[tex]\hat p=0.38[/tex] estimated proportion of interesst
[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)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the proportion is lower than 0.5:
Null hypothesis:[tex]p \geq 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].
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.38 -0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}=-2.4[/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 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<-2.4)=0.0082[/tex]
And we can use the following excel code to find it:
"=NORM.DIST(-2.4,0,1,TRUE)"
