Respuesta :
Answer:
95% confidence interval for p is [0.54 , 0.75].
Step-by-step explanation:
We are given that a police officer would like to estimate the proportion of petty theft cases that have gone unsolved in a large city. He takes a random sample of 81 petty theft cases and finds 52 of these have gone unsolved.
Let p = population proportion of petty theft cases
So, the pivotal quantity for 95% confidence interval for population proportion p is given by;
      P.Q. = [tex]\frac{\hat p -p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = proportion of petty theft cases that have gone unsolved in a large city in a sample of 81 cases = [tex]\frac{52}{81}[/tex]
      n = sample of cases = 81
So, 95% confidence interval for the population proportion, p is ;
P(-1.96 < N(0,1) < 1.96) = 0.95
P(-1.96 < [tex]\frac{\hat p -p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } }[/tex] < 1.96) = 0.95
P( [tex]-1.96 \times {\sqrt{\frac{\hat p(1- \hat p)}{n} } }[/tex] < [tex]{\hat p -p}[/tex] < [tex]1.96 \times {\sqrt{\frac{\hat p(1- \hat p)}{n} } }[/tex] ) = 0.95
P( [tex]\hat p -1.96 \times {\sqrt{\frac{\hat p(1- \hat p)}{n} } }[/tex] < p < [tex]\hat p +1.96 \times {\sqrt{\frac{\hat p(1- \hat p)}{n} } }[/tex] ) = 0.95
95% confidence interval for p = [ [tex]\hat p -1.96 \times {\sqrt{\frac{\hat p(1- \hat p)}{n} } }[/tex] , [tex]\hat p +1.96 \times {\sqrt{\frac{\hat p(1- \hat p)}{n} } }[/tex] ]
                       = [ [tex]\frac{52}{81} -1.96 \times {\sqrt{\frac{\frac{52}{81} (1- \frac{52}{81} )}{81} } }[/tex] , [tex]\frac{52}{81} +1.96 \times {\sqrt{\frac{\frac{52}{81} (1- \frac{52}{81} )}{81} } }[/tex] ]
                       = [0.54 , 0.75]
Therefore, 95% confidence interval for p is [0.54 , 0.75].
