As a condition of employment, Fashion Industries applicants must pass a drug test. Of the last 240 applicants, 26 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 11% of the applicants fail the test?   NoYes

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Asked Nov 18, 2019
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As a condition of employment, Fashion Industries applicants must pass a drug test. Of the last 240 applicants, 26 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 11% of the applicants fail the test?
   
 
  • No
  • Yes
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Expert Answer

Step 1

a.

The 90% confidence interval for the proportion of applicants that fail the test is,

Step-by-step procedure to obtain the value of  using StatKey:

 

  • Choose Normal under Theoretical Distributions.
  • In Normal Distribution, click Edit Parameters.
  • Enter mean as 0 and standard deviation as 1 and click Ok.
  • Click the Centre Proportion value in graph and enter 90.
  • Click Two-Tail.
  • Click OK.

Output using the StatKey is given below:

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Normal Distribution Reset Plot Left Tail Two Tail Right Tail 0.40 0.35 0.30 0.25 0.20 0.900 0.050 0.050 0.15 0.10 0.05 0.00 -4.0 0.0 -3.0 -2.0 1.0 1.0 2.0 3.0 4.0 -1.645 1.645

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Step 2

From the output, the value of  is 1.645 for 90% confidence.

The value of  p^ is obtained as follows:

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26 240 =0.1083

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Step 3

The 90% confidence interval i...

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p(1-p) Sample statistic tz* SE = p±z*. 0.1083(1-0.1083) 0.1083 t1.645, 240 =0.1083 t 0.033 = (0.0753,0.1413)

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