Answer:
95% confidence interval to estimate the true proportion of evens rolled on a die.
(0.197368 , 0.762632)
Step-by-step explanation:
Step-by-step explanation:-
Given A fit is rolled 25 times and 12 evens are observed
proportion
![p = (x)/(n) = (12)/(25) = 0.48](https://img.qammunity.org/2022/formulas/mathematics/high-school/7n2d25p1jswziibgp258stisgjvavk438c.png)
q = 1 - p = 1- 0.48 = 0.52
Level of significance =0.05
![Z_(0.05) = 1.96](https://img.qammunity.org/2022/formulas/mathematics/high-school/p0ixbzgyv81a0m874tjogp7yltocwcn7jl.png)
95% confidence interval to estimate the true proportion of evens rolled on a die.
![(p^(-) - Z_(0.05) \sqrt{(p(1-p))/(n) } , p^(-) + Z_(0.05) \sqrt{(p(1-p))/(n) } )](https://img.qammunity.org/2022/formulas/mathematics/high-school/ck7gdlmuv7am2v58ng0lup73t1mdmsegki.png)
![(0.48 - 1.96 \sqrt{(0.48(1-0.48))/(12) } , 0.48 + 1.96 \sqrt{(0.48(1-0.48))/(12) } )](https://img.qammunity.org/2022/formulas/mathematics/high-school/qrv4tsdtqjo53yyyla001958oil1lhe0fn.png)
( 0.48 - 1.96 (0.1442 , 0.48 + 1.96(0.1442)
( 0.48 - 0.282632 , 0.48 + 0.282632)
(0.197368 , 0.762632)
Final answer:-
95% confidence interval to estimate the true proportion of evens rolled on a die.
(0.197368 , 0.762632)