Respuesta :
Answer:
As the P-value (0.086) is greater than the significance level (0.01), the effect is not significant.
The null hypothesis failed to be rejected.
There is not enough evidence to support the claim that the return rate is less than 15%.
Step-by-step explanation:
This is a hypothesis test for a proportion.
The claim is that the return rate is less than 15%.
Then, the null and alternative hypothesis are:
[tex]H_0: \pi=0.15\\\\H_a:\pi<0.15[/tex]
The significance level is 0.01.
The sample has a size n=5000.
The sample proportion is p=0.143.
[tex]p=X/n=717/5000=0.143[/tex]
The standard error of the proportion is:
[tex]\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.15*0.85}{5000}}\\\\\\ \sigma_p=\sqrt{0.000026}=0.005[/tex]
Then, we can calculate the z-statistic as:
[tex]z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.143-0.15+0.5/5000}{0.005}=\dfrac{-0.007}{0.005}=-1.366[/tex]
This test is a left-tailed test, so the P-value for this test is calculated as:
[tex]P-value=P(z<-1.366)=0.086[/tex]
Answer:
[tex]z=\frac{0.1434 -0.15}{\sqrt{\frac{0.15(1-0.15)}{5000}}}=-1.307[/tex]
Now we can calculate the p value with the following probability:
[tex]p_v =P(z<-1.307)=0.0956[/tex]
For this case the p value is lower than the significance level provided of 0.1 so then we can reject the null hypothesis and we can conclude that the the return rate is significantly less than 15%.
Step-by-step explanation:
Information provided
n=5000 represent the random sample taken
X=717 represent the surveys returned
[tex]\hat p=\frac{717}{5000}=0.1434[/tex] estimated proportion of urvyes returned
[tex]p_o=0.15[/tex] is the value to verify
[tex]\alpha=0.01[/tex] represent the significance level
z would represent the statistic
[tex]p_v[/tex] represent the p value
System of hypothesis
We want to verify if the return rate is less than 15% the system of hypothesis are.:
Null hypothesis:[tex]p \geq 0.15[/tex]
Alternative hypothesis:[tex]p < 0.15[/tex]
The statistic is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
Replacing the info we got:
[tex]z=\frac{0.1434 -0.15}{\sqrt{\frac{0.15(1-0.15)}{5000}}}=-1.307[/tex]
Now we can calculate the p value with the following probability:
[tex]p_v =P(z<-1.307)=0.0956[/tex]
For this case the p value is lower than the significance level provided of 0.1 so then we can reject the null hypothesis and we can conclude that the the return rate is significantly less than 15%.
