Question

As a condition of employment, Fashion Industries applicants must pass a drug test. Of the last 220 applicants, 18 failed the test. (Use z Distribution Table.) a. Develop a 90% confidence interval for the proportion of applicants that fail the test. (Round your answers to 3 decimal places.) For the applicants the confidence interval is between and . b. Would it be reasonable to conclude that more than 12% of the applicants fail the test?

Answer 1

Solution:

a)

Given,

n = 220 ....... Sample size

x = 18 .......no. of successes in the sample

Let $\hat p$ denotes the sample proportion.

  $\therefore$  $\hat p$ = x/n = 18/220 = 0.0818

Our aim is to construct 90% confidence interval.

$\therefore$ c = 0.90

$\therefore$$\alpha$ = 1- c = 1- 0.90 = 0.10

$\therefore$  $\alpha$/2 = 0.10 $\slash$ 2 = 0.05 and 1- $\alpha$ /2 = 0.950

$\therefore$ = 1.645 (use z table)

2012

Now , the margin of error is given by

E =
2012
*  
Vp(1-P)/n

= 1.645 * $url=https://latex.codecogs.com/gif.latex?%5Csqrt%7B%7D&t=1596245000&ymreqid=f4764063-0d40-e2b0-1cde-f0029e012f00&sig=70dcK_NMT7oJ15MinO1o_g--~D$ [0.0818 *(1 - 0.0818)/220]

= 0.0304

Now the confidence interval is given by

($\hat p$ - E)   ($\hat p$ + E)

(0.0818   - 0.0304)   (0.0818   + 0.0304)

0.051   0.112

For the applicants the confidence interval is between 0.051 and  0.112

b)

Claim : More than 12% of the applicants fail the test.

i.e. p > 0.12

Hypothetical value 0.12 is not in the interval. But lower limit 0.051 is not greater than 0.12

So , fail to reject H₀ : p $\leq$ 0.12

Answer is

No , it would not be reasonable to conclude that more than 12% of the applicants fail the test.