POW_Package Delivery by Drones

.xlsx

School

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.
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