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 15% 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, 85% 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?

___________ orders

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

( , )

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

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

( , )

(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.)

( , )

Percents

( , )

Explanation / Answer

Q1.
Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Sample Size(n)=200
Sample proportion = x/n =0.15
Confidence Interval = [ 0.15 ±Z a/2 ( Sqrt ( 0.15*0.85) /200)]
= [ 0.15 - 1.645* Sqrt(0.001) , 0.15 + 1.65* Sqrt(0.001) ]
= [ 0.108,0.192]
Q2.
Sample Size(n)=200
Sample proportion = x/n =0.85
Confidence Interval = [ 0.85 ±Z a/2 ( Sqrt ( 0.85*0.15) /200)]
= [ 0.85 - 1.645* Sqrt(0.001) , 0.85 + 1.65* Sqrt(0.001) ]
= [ 0.808,0.892]
Q3.
Confidence Interval for Diffrence of Proportion
CI = (p1 - p2) ± Z a/2 Sqrt(p1(1-p1)/n1 + p2(1-p2)/n2 )
Proportion 1
No. of chances( X1 )=30
No.Of Observed (n1)=200
P1= X1/n1=0.15
Proportion 2
No. of chances(X2)=170
No.Of Observed (n2)=200
P2= X2/n2=0.85
C.I = (0.15-0.85) ±Z a/2 * Sqrt( (0.15*0.85/200) + (0.85*0.15/200) )
=(0.15-0.85) ± 1.64* Sqrt(0.001)
=-0.7-0.059,-0.7+0.059
=[-0.759,-0.641]

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