Question

A fit is rolled 25 times and 12 evens are observed. Calculate and interpret a 95% confidence interval to estimate the true proportion of evens rolled on a die.

Answer 1

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    $p = \frac{x}{n} = \frac{12}{25} = 0.48$

               q = 1 - p = 1- 0.48 = 0.52

Level of significance =0.05

$Z_{0.05} = 1.96$

95% confidence interval to estimate the true proportion of evens rolled on a die.

$(p^{-} - Z_{0.05} \sqrt{\frac{p(1-p)}{n} } , p^{-} + Z_{0.05} \sqrt{\frac{p(1-p)}{n} } )$

$(0.48 - 1.96 \sqrt{\frac{0.48(1-0.48)}{12} } , 0.48 + 1.96 \sqrt{\frac{0.48(1-0.48)}{12} } )$

( 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)