Answer: (0.3751,0.4821)
Explanation:
Given : Level of significance :
![1-\alpha:0.95](https://img.qammunity.org/2020/formulas/mathematics/college/iovh9ozk91mzido8mb90kgcvgunj0ml0cw.png)
Then , significance level :
![\alpha: 1-0.95=0.05](https://img.qammunity.org/2020/formulas/mathematics/high-school/9x6075632zgcvqcj0z3yy9jc9lp14p66n9.png)
Since , sample size :
, i.e. a large sample (n<30).
Then we use z-test.
Using excel (by going in formulas for more statistics and then statistics), Critical value :
![z_(\alpha/2)=1.96](https://img.qammunity.org/2020/formulas/mathematics/high-school/fn1e1isyr7r4ubq2yxfnpgs4mo3eo8m7ik.png)
Also, the proportion of people said that they were fans of the visiting team :-
![\hat{p}=( 141)/(329)\approx 0.4286](https://img.qammunity.org/2020/formulas/mathematics/college/skn0n47i3r2rm172vtwza871ulx81qemyz.png)
The confidence interval for population proportion is given by :-
![\hat{p}\pm z_(\alpha/2)\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}](https://img.qammunity.org/2020/formulas/mathematics/college/pxj5zva1u6igd7xybc6x9ei113i3mgmyt5.png)
![0.4286\pm(1.96)\sqrt{(0.4286(1-0.4286))/(329)}\\\\\approx0.4286\pm0.053\\\\=(0.4286-0.0534, 0.4286+0.053=(0.3751,\ 0.4821)](https://img.qammunity.org/2020/formulas/mathematics/college/oo9n13nua300vw5emt16t5mk8e5842qqks.png)
Hence, a 95% confidence interval for the population proportion of attendees who were fans of the visiting team= (0.3751,0.4821)