Respuesta :
Answer:
[tex]z=\frac{0.185 -0.2}{\sqrt{\frac{0.2(1-0.2)}{6997}}}=-3.137[/tex]
[tex]p_v =P(z<-3.137)=0.0009[/tex]
Since the p value is very low we have enough evidence to reject the null hypothesis at the significance lvel given of 0.01 so then we can conclude that the return rate ofr this case is significantly lower than 20%
Step-by-step explanation:
Data given
n=6997 represent the random sample taken
X=1296 represent the returned surveys
[tex]\hat p=\frac{1296}{6997}=0.185[/tex] estimated proportion of returned surveys
[tex]p_o=0.2[/tex] is the value that we want to test
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
System of hypothesis
We need to conduct a hypothesis in order to test the claim that the true proportion of returned surveys is less than 0.2.:
Null hypothesis:[tex]p\geq 0.2[/tex]
Alternative hypothesis:[tex]p < 0.2[/tex]
The statistic is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
Calculate the statistic
Since we have all the info required we can replace in formula (1) like this:
[tex]z=\frac{0.185 -0.2}{\sqrt{\frac{0.2(1-0.2)}{6997}}}=-3.137[/tex]
Statistical decision
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<-3.137)=0.0009[/tex]
Since the p value is very low we have enough evidence to reject the null hypothesis at the significance lvel given of 0.01 so then we can conclude that the return rate ofr this case is significantly lower than 20%
