Question

A random sample of 10 houses in Loonah, each of which is heated with natural machining, is selected and the amount of gas (safe gas) used during the month of Feb is determined for each house. The resulting observations are 104, 113, 114, 127, 112, 115, 124, 123 161, 120. Assuming that these were selected from a normal population, calculate:

a.) a point estimate for the mean (ie. )

b.) an unbiased point estimate for the standard deviation (ie. S) (Show equation for calculating.)

c.) Calculate a 90% Confidence Interval for your answer in part a using the information in part b.

d.) Calculate a 90% Prediction Interval for the next house have safe gas indicated by.

e.) Calculate a Tolerance Interval that has a 95% Confidence of capturing at least 90% of the values in a normal population.

Answer 1

a)

Sample Mean,    x̅ = Σx/n = 121.3

b)

X	(X - X̄)²
104	299.290
113	68.890
114	53.290
127	32.490
112	86.490
115	39.690
124	7.290
123	2.9
161	1576.090
120	1.690

	X	(X - X̄)²
total sum	1213	2168.100
n	10	10

sample std dev =   √ [ Σ(X - X̄)²/(n-1)] =   15.5210

c)

Level of Significance ,    α =    0.05
sample std dev ,    s =    15.5210
Sample Size ,   n =    10
Sample Mean,    x̅ =   121.3000

degree of freedom=   DF=n-1=   9      
't value='   tα/2=   2.2622   [Excel formula =t.inv(α/2,df) ]  
              
Standard Error , SE =   s/√n =   4.9082      
margin of error ,   E=t*SE =   11.103      

confidence interval is               
Interval Lower Limit=   x̅ - E =    110.1970      
Interval Upper Limit=   x̅ + E =    132.4030      
confidence interval is (   110.1970   < µ <   132.4030   )

d)

margin of error for prediction interval,E=   t*s*√(1+1/n)=   36.8246

   
prediction interval is               
Interval Lower Limit=   x̅ - E =    84.4754      
Interval Upper Limit=   x̅ + E =    158.1246  

  
prediction interval is (   84.4754   < µ <   158.1246   )