Respuesta :
Answer: (0.3409, 0.4459)
Step-by-step explanation:
Given : Level of significance : [tex]1-\alpha:0.95[/tex]
Then , significance level : [tex]\alpha: 1-0.95=0.05[/tex]
Since , sample size : [tex]n=333[/tex]
Using excel (by going in more formulas and then statistics), Critical value : [tex]z_{\alpha/2}=1.96[/tex]
Also, the proportion of people said that they were fans of the visiting team :-
[tex]\hat{p}=\dfrac{ 131}{333}\approx0.3934[/tex]
The confidence interval for population proportion is given by :-
[tex]\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
[tex]0.3934\pm(1.96)\sqrt{\dfrac{0.3934(1-0.3934)}{333}}\\\\\approx0.3934\pm0.0525\\\\=(0.3934-0.0525, 0.3934+0.0525=(0.3409,\ 0.4459) [/tex]
Hence, a 95% confidence interval for the proportion of employees who have a daily commute longer than 30 minutes= (0.3409, 0.4459)