Question

The Occupational Safety & Health Administration requires companies that handle hazardous chemicals to complete material safety data sheets (MSDS). These MSDS have been criticized for being too hard to understand and complete by workers. A study of 150 MSDS revealed that only 12% were satisfactorily completed. (Chemical & Engineering News, Feb. 7, 2005.)

Give a 98% confidence interval for the true percentage of MSDS that are satisfactorily completed.

Answer 1

98% confidence interval for the true percentage of MSDS that are satisfactorily completed is [0.058 , 0.182].

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. =
`\frac{\hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` ~ N(0,1)

where,
`\hat p` = 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 <
`\frac{\hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` < 2.33) = 0.98

P(
`-2.33 * {\sqrt{(\hat p(1-\hat p))/(n) } }` <
`{\hat p-p}` <
`2.33 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ) = 0.98

P(
`\hat p-2.33 * {\sqrt{(\hat p(1-\hat p))/(n) } }` < p <
`\hat p+2.33 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ) = 0.98

98% confidence interval for p =

[
`\hat p-2.33 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ,
`\hat p+2.33 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ]

= [
`0.12-2.33 * {\sqrt{(0.12(1-0.12))/(150) } }` ,
`0.12+2.33 * {\sqrt{(0.12(1-0.12))/(150) } }` ]

= [0.058 , 0.182]

Therefore, 98% confidence interval for the true percentage of MSDS that are satisfactorily completed is [0.058 , 0.182].