Respuesta :
Answer:
a) [tex]z=\frac{0.4704 -0.45}{\sqrt{\frac{0.45(1-0.45)}{1250}}}=1.450[/tex]
Now we can calculate the p value given by:
[tex]p_v =P(z>1.450)=0.0735[/tex]
Since the p value is higher than the significance level of [tex]\alpha=0.05[/tex] we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true proportion is significantly higher than one year ago (0.45)
b) [tex] 0.4704 -1.96 \sqrt{\frac{0.4704*(1-0.4704)}{1250}}= 0.443[/tex]
[tex] 0.4704 +1.96 \sqrt{\frac{0.4704*(1-0.4704)}{1250}}= 0.498[/tex]
Step-by-step explanation:
Information given
n=1250 represent the sample size selected
X=588 represent the people who participate in recycling programs.
[tex]\hat p=\frac{588}{1250}=0.4704[/tex] estimated proportion of people who participate in recycling programs.
[tex]p_o=0.45[/tex] is the value to compare
[tex]\alpha=0.05[/tex] represent the significance level
z would represent the statistic
[tex]p_v[/tex] represent the p value
Part a
We want to test the hypothesis that the proportion of the population who participate in recycling programs is greater than it was one year ago (0.45):
Null hypothesis:[tex]p\leq 0.45[/tex]
Alternative hypothesis:[tex]p > 0.45[/tex]
The statistic would be given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
Replacing the info given we got:
[tex]z=\frac{0.4704 -0.45}{\sqrt{\frac{0.45(1-0.45)}{1250}}}=1.450[/tex]
Now we can calculate the p value given by:
[tex]p_v =P(z>1.450)=0.0735[/tex]
Since the p value is higher than the significance level of [tex]\alpha=0.05[/tex] we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true proportion is significantly higher than one year ago (0.45)
Part b
the confidence interval would be given by:
[tex] \hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
And the critical value for a 95% confidence interval can be obtained in the normal distribution and we got [tex] z_{\alpha/2} =1.96[/tex]. And replacing into the confidence interval formula we got:
[tex] 0.4704 -1.96 \sqrt{\frac{0.4704*(1-0.4704)}{1250}}= 0.443[/tex]
[tex] 0.4704 +1.96 \sqrt{\frac{0.4704*(1-0.4704)}{1250}}= 0.498[/tex]
