1. How well materials conduct heat matters when designing houses, f...

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Question 1:

a. Using the probability plot and a box plots it reasonable to use the t procedures? Why or why not?
Using the below two plots we can say it is reasonable to use the t procedures because the normal probability plots shows that points close to the straight line and inside the 95% confidence limits. Also the standard deviation is very small in this case, It means data is normally distributed. The p-value associated with normal probability plot is 0.209 which is bigger than 0.05 so we can conclude that the data for heat conductivity is normally distributed. The same can also be seen from the box plot where we can see that the data is equally spread without any outliers. These both plots show that the assumption of using the t procedures are met and we can use the t distribution on this data.

b. If your conclusion in part (a) is yes, give a 95% confidence interval for the mean conductivity. (Make sure your interpretation of the 95% confidence interval is in a complete sentence in the context of the problem.)
95% confidence interval for mean conductivity is given below:
Lower Limit = Sample Mean – Critical Value of t at n-1 degree of freedom*Standard Error
Upper Limit = Sample Mean + Critical Value of t at n-1 degree of freedom*Standard Error
Here we have,
Sample mean = 1.1155
Sample standard deviation = 0.04034 and n = 11
Standard Error = Sample standard deviation/sqrt(n) = 0.04034/sqrt(11) =0.012162968
Critical value at 5% level of significance with n-1=11-1=10 degree of freedom is ±2.22813
Lower Limit = 1.1155 - 2.22813*0.012162968 = 1.088399
Upper Limit = 1.1155 + 2.22813*0.012162968 = 1.142601
We are 95% confident that population mean heat conductivity of a glass lies between 1.088399 and 1.142601....

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