Respuesta :
Answer:
98% confidence interval for the true percentage of MSDS that are satisfactorily completed is [0.058 , 0.182].
Step-by-step explanation:
We are given that a study of 150 MSDS revealed that only 12% were satisfactorily completed.
Firstly, the Pivotal quantity for 98% confidence interval for the population proportion 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] = sample proportion of MSDS that were satisfactorily completed = 12%
n = sample of MSDS = 150
p = true percentage of MSDS
Here for constructing 98% confidence interval we have used One-sample z test for proportions.
So, 98% confidence interval for the true percentage, p is ;
P(-2.33 < N(0,1) < 2.33) = 0.98 {As the critical value of z at 1% level
of significance are -2.33 & 2.33}
P(-2.33 < [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < 2.33) = 0.98
P( [tex]-2.33 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < [tex]{\hat p-p}[/tex] < [tex]2.33 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.98
P( [tex]\hat p-2.33 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < p < [tex]\hat p+2.33 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.98
98% confidence interval for p =
[ [tex]\hat p-2.33 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] , [tex]\hat p+2.33 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ]
= [ [tex]0.12-2.33 \times {\sqrt{\frac{0.12(1-0.12)}{150} } }[/tex] , [tex]0.12+2.33 \times {\sqrt{\frac{0.12(1-0.12)}{150} } }[/tex] ]
= [0.058 , 0.182]
Therefore, 98% confidence interval for the true percentage of MSDS that are satisfactorily completed is [0.058 , 0.182].
