Question

1. Suppose you sell vacuums. Suppose also you know the lifespan of your vacuums (the length of time from purchase to landfill) to be given by independent identically distributed normal random variables, but you don't know what the average lifespan is. If you’re allowed to follow up with 20 customers and find out the lifespans of their vacuums, how would you use that information to give your best guess of what the actual average lifespan is? Bear in mind that you have to give a single number as your best guess. In what sense is your guess a good guess?

2. Suppose that a manufacturer claims the average mass of the bolts they produce is 1.5g. Your boss decides to test this by buying 100 such bolts and finding the average mass of those 100. The boss decides they’ll believe the manufacturer’s claim if the average of those 100 falls within 0.01g of 1.5 g. Find the significance level, α, of this test. Give a numerical value.

What if your boss decides to make the acceptance region larger, will this make α Go up or down ? Explain why. What about β? Wil β get larger or smaller if the acceptance region gets larger? Explain why.

Please answer both questions with as much work as possible

thank you

Answer 1

Solution :

(a)

sample mean is unbiased estimator of population mean

average life-span will be the average of life-span of 20 customers

Let the life-spans be l1,l2,...,l20

Average life span = 1/20 * (l1 + l2 + l3 + l4 + ... l18 + l19 + l20)

This is a good guess since you are following 20 customers, which are coming randomly in the shop, they will provide proper random distribution of bigger chunk of data. So, in that sense our guess is a good guess

b)

Standard Deviation , σ =    0.1                  
sampling error , E =   0.01                  
Confidence Level , CL=   68%                  
                          
alpha = 1-CL = 32%                  
Z value =    Zα/2 =    0.999   [excel formula =normsinv(α/2)]              
                          
Sample Size,n = (Z * σ / E )² = ( 0.999 * 0.1 / 0.01 ) ² =  99.715

So,Sample Size needed = 100                  

alpha = 0.32

Alpha go up because confidence interval decraese for large acceptance area

Beta will go down.

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