Answer to Question #223277 in Statistics and Probability for Lucky

Question

For the production process of ballpoint pens of a certain type, it is known that 5% of the pens are

defective. The manager believes that this percentage is too high and hence changes the production

process. To test the merits of this new process, a random sample of 350 pens is taken; only 9 of

them are defective.

(a) (i) Determine a 90% confidence interval for p, the population proportion of defective pens

produced with the new process. (6)

(ii) Interpret the confidence interval. (2)

(iii) Do you think that the new process is better? (1)

Answer 1

(a).

i. $90\% \space confidence \space interval$

$CI=\hat P\pm z.(\sqrt{\frac{\hat P(1-\hat P)}{n}})$

$z=1.645$

$\hat P=\frac{x}{n}=\frac{9}{350}=0.026\\$

$where \space x=9 \space (we \space use \space new \space value \space of \space defective \space , \space old \space value \space of \space defective \space is \space \\ 5 \space \% \space dont \space use \space in \space calculation \space of \space new-process)\\\space n=350$

$CI=0.026\pm1.645(\sqrt{\frac{\frac{9}{350}(1-\frac{9}{350})}{350}})$

$=(0.012, 0.04)$

$0.012\le p\le0.04)$

ii.

$Therefore, \space based \space on \space the \space data \space provided \space , \space the \space \space 90\% \\ \space confidence \space interval \space for \space the \space population \space \space proportion \space is \space \\$

$0.012\le P\le0.04)$

$which \space indicates \space that \space we \space are \space 90 \space \%$

$confident \space that \space the \space true \space population \space proportion \\ \space p \space is \space contained \space by \space the \space interval (0.012,0.04).$

iii.

$The \space new \space \space process \space is \space better \space since \space it \space lowers \space proportion \space \space of \\ \space defective \space pens \space significantly \space .$