The Ping Company makes custom-built golf clubs and competes in the $4 billion go

Question

The Ping Company makes custom-built golf clubs and competes in the $4 billion golf equipment industry. To improve its business processes, Ping decided to seek ISO 9001 certification. As part of this process, a study of the time it took to repair golf clubs sent to the company by mail determined that 10% of orders were sent back to the customers in 5 days or less. Ping examined the processing of repair orders and made changes. Following the changes, 88% of orders were completed within 5 days. Assume that each of the estimated percents is based on a random sample of 200 orders.

(a) How many orders were completed in 5 days or less before the changes?

180 Incorrect: Your answer is incorrect. orders

Give a 90% confidence interval for the proportion of orders completed in this time. (Round your answers to three decimal places.)

173 Incorrect: Your answer is incorrect. , 187 Incorrect: Your answer is incorrect.

(b) How many orders were completed in 5 days or less after the changes?

176 Correct: Your answer is correct. orders

Give a 90% confidence interval for the proportion of orders completed in this time. (Round your answers to three decimal places.)

168.417 Incorrect: Your answer is incorrect. , 183.583 Incorrect: Your answer is incorrect.

(c) Give a 90% confidence interval for the improvement. Express this both for a difference in proportions and for a difference in percents. (Define the groups so that the difference will be positive. Round your answers for proportions to three decimal places and round your answers for percents to one decimal place.)

proportions 3.30 Incorrect: Your answer is incorrect. , 11.30 Incorrect: Your answer is incorrect.

percents ,

Explanation / Answer

a) as 10% orders were sent back,

hence number of order completed in 5 days or less before the changes =np=0.1*200 =20

here std error =(p(1-p)/n)1/2 where p=0.1 ; n=200

   =0.0212

also for 90% CI, z =1.6449

hence confidence interval =mean proportion +/- z*std error =0.065 ; 0.135

b) number of order to be completed =np=200*0.88 =176

here std error =

(p(1-p)/n)1/2 where p=0.88 ; n=200

   =0.023

also for 90% CI, z =1.6449

hence confidence interval =mean proportion +/- z*std error =0.842 ; 0.918

c) here p1=0.88 ; p2 =0.1 and n1=n2 =200

hence std error =((p1(1-p1)/n1 +p2(1-p2)/n2)1/2 =0.031

therefore confidence interval =(p1-p2)+/- z*std error =0.729 ; 0.831

or 72.9% to 83.1%

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