Respuesta :
Answer:
95% confidence interval to estimate the true proportion of evens rolled on a die.
(0.197368 , 0.762632)
Step-by-step explanation:
Explanation:-
Given A fit is rolled 25 times and 12 evens are observed
proportion [tex]p = \frac{x}{n} = \frac{12}{25} = 0.48[/tex]
q = 1 - p = 1- 0.48 = 0.52
Level of significance =0.05
[tex]Z_{0.05} = 1.96[/tex]
95% confidence interval to estimate the true proportion of evens rolled on a die.
[tex](p^{-} - Z_{0.05} \sqrt{\frac{p(1-p)}{n} } , p^{-} + Z_{0.05} \sqrt{\frac{p(1-p)}{n} } )[/tex]
[tex](0.48 - 1.96 \sqrt{\frac{0.48(1-0.48)}{12} } , 0.48 + 1.96 \sqrt{\frac{0.48(1-0.48)}{12} } )[/tex]
( 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)
