Respuesta :
Answer:
[tex]z=\frac{0.55 -0.6}{\sqrt{\frac{0.6(1-0.6)}{80}}}=-0.913[/tex]
[tex]p_v =P(z<-0.913)=0.181[/tex]
So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we see that [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 5% of significance the proportion of graduate students show that only 44 students have ever done so is not significantly lower than 0.6
Step-by-step explanation:
Data given and notation
n=80 represent the random sample taken
X=44 represent the students that have bought merchandise on-line at their site
[tex]\hat p=\frac{44}{80}=0.55[/tex] estimated proportion of graduate students show that only 44 students have ever done so
[tex]p_o=0.6[/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 true proportion of interest is lower than 0.6 or 60%, so then the system of hypothesis are.:
Null hypothesis:[tex]p \geq 0.6[/tex]
Alternative hypothesis:[tex]p < 0.6[/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{0.55 -0.6}{\sqrt{\frac{0.6(1-0.6)}{80}}}=-0.913[/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 for this case is [tex]\alpha=0.05[/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<-0.913)=0.181[/tex]
So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we see that [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 5% of significance the proportion of graduate students show that only 44 students have ever done so is not significantly lower than 0.6
