POW_Package Delivery by Drones
xlsx
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College of San Mateo *
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Course
123
Subject
Industrial Engineering
Date
May 25, 2024
Type
xlsx
Pages
3
Uploaded by GeneralWillpowerAntelope75
Package Delivery by Drones. Amazon.com is testing the use of drone
delivery times must be sufficiently small. Consider a sample of 24 dro
On the basis of above information, answer the following questions:
Question 1. Construct a 90% confidence interval estimate of the pop
Hint:
Step1: Calculate the degrees of freedom and alpha coefficient for 90%
Step 2: Calculate chi-square values for left tail and right tail by either u
1-alpha/2) or
use CHISQ.INV (alpha/2, degrees of freedom) and CHISQ
Step 3: Calculate the confidence interval of variance by multiplying d
Question 2. Construct a 90% confidence interval estimate of the pop
Hint: Take the square root of the confidence limits of the variance
Answer: 1. The confidence interval estimate of the population variance is 0.52
First find the degrees of freedom, alpha and alpha/2 which is 23 , 0.1 Then calculate the left tail using =CHISQ.INV(0.05, 23) = 13.09051 an
23) = 35.17246 To find the confidence interval lower limit ( 23 * 0.81) / 35.17246 = 0.
* 0.81) / 13.09051 = 1.423168 2. The confidence interval estimate of the population standard deviati
To get these values, square root the values of the confidence interval
sample size 24
sample variance 0.81
degrees of freedom 23
Confidence 0.9
alpha
0.1
alpha/2
0.05
left tail 13.090514
right tail
35.172462
lower limit
0.5296758
upper limit
1.4231679
sqrt(lower limit)
0.7277883
sqrt(upper limit)
1.192966
es to deliver packages for same-day delivery. In order to quote narrow time wind
one deliveries with sample variance of s
2
=0.81 pulation variance for the drone delivery time.
% confidence, Then calculate alpha/2
using the Chi-Square Table Page 493 of the textbook (use degrees of freedom a
Q.INV.RT (alpha/2, degrees of freedom) if using Excel
degrees of freedom times sample variance and then divide by the lower tail an
pulation standard deviation.
29676 <= σ^2<= 1.423168 , 0.005
nd lright tail using =CHISQ.INV.RT(0.05, .529676, and for the upper limit ( 23 tion is 0.727788 <=σ <= 1.192966
estimate of the population variance.
dows, the variability in and alpha/2 and nd upper tail chi-square values
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