Question

QUESTION 2 (2.5 + 2 + 3.5 + 2 = 10 marks)

The length of time before roof tiles need replacing has a mean of 20 years and a standard deviation of 6 years. The length of time until a roof tile needs replacement is also found to have an approximately normal distribution.

1. What is the proportion of roof tiles that last for more than 22 years? Show all working, define the variable, state the distribution and give your answer to 3 decimal places.

2. What is the value of the 20^th percentile for roof tiles before they need replacing? Give your answer correct to one decimal place only. Show all working. A diagram may also be helpful.

3. Suppose we do not know the true mean length of time all roof tiles last. We take a sample of 40 pieces of Super roof tiles and the mean is found to be 19 years with a standard deviation of 5.8 years. Find a 95% confidence interval for the mean lifespan of all Super roof tiles. Show all working and interpret the interval estimate.

4. We calculated the 99% confidence interval for the mean lifespan of all Super roof tiles and obtained the following answers:

   Lower Limit:16.52 years
   Upper Limit: 21.48 years

   Based on the information in (c) and (d), do the 95% and 99% confidence intervals contain the true mean lifespan of all roof tiles? Comment on your results in relation to the mean lifespan of Super roof tiles and across all roof tiles. What does the evidence tell us?

Answer 1

a)

µ =    20                  
σ =    6                  
                      
P ( X ≥   22   ) = P( (X-µ)/σ ≥ (22-20) / 6)              
= P(Z ≥   0.33   ) = P( Z <   -0.333   ) =    0.3694 (answer)
....

b)

µ=   20                  
σ =    6                  
P(X≤x) =   0.2                  
                      
Z value at    0.2   =   -0.8416   (excel formula =NORMSINV(   0.2   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   -0.842   *   6   +   20  
X   =   15.0   (answer)          
...........

c)

sample std dev ,    s =    5.8000
Sample Size ,   n =    40
Sample Mean,    x̅ =   19.0000

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   39          
't value='   tα/2=   2.023   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   5.8/√40=   0.9171          
margin of error , E=t*SE =   2.0227   *   0.9171   =   1.8549
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    19.00   -   1.8549   =   17.1451
Interval Upper Limit = x̅ + E =    19.00   -   1.8549   =   20.8549
95%   confidence interval is (   17.1   < µ <   20.9   )
There is 95% confidence that true mean lies within confidence interval
...

d)

95% and 99% confidence intervals contain the true mean lifespan of all roof tiles

............

thanks

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