Question

The data below gives the mean price (in cents) of a litre of regular gasoline at self-service filling stations at a sample of six urban centres in Canada in May 2013.

Saint John, NB...........122.7

Quebec.........132.6

Toronto.............125.4

Regina.......123.7

Saskatoon...........125.2

Edmonton.........114

Provide answers to the following to two decimal places.

Part (a) Find the sample mean (in cents).  

Part (b) Find the sample standard deviation (in cents).  

Part (c) Using the appropriate t distribution, find a 90% confidence interval for the mean gasoline price per litre across urban Canada in May 2013. Provide the upper and lower bounds for your confidence interval. (  ,  )

Part (d) Would a 95% confidence interval be wider or narrower than the interval you found in Part (c)? Using the appropriate t distribution, find a 95% confidence interval for the mean gasoline price per litre across urban Canada in May 2013. Provide the upper and lower bounds for your confidence interval. (  ,  )

Answer 1

X	(X - X̄)²
122.7	1.521
132.6	75.111
125.4	2.151
123.7	0.054
125.2	1.604
114	98.671

	X	(X - X̄)²
total sum	743.6	179.113
n	6	6

a)

mean =    ΣX/n =    743.600   /   6.000   =   123.93

b)

sample variance =    Σ(X - X̄)²/(n-1)=   179.1133   /   5.0000   =   35.8227
                      
sample std dev =   √ [ Σ(X - X̄)²/(n-1)] =   √   35.8227   =       5.99

c)

sample std dev ,    s = 5.99
Sample Size ,   n =    6
Sample Mean,    x̅ = ΣX/n =    123.93

Level of Significance ,    α =    0.10   
degree of freedom=   DF=n-1=   5          
't value='   tα/2=   2.0150   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   5.985   / √   6   =   2.4434
margin of error , E=t*SE =   2.0150   *   2.4434   =   4.924
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    123.93 -   4.924   =   119.010
Interval Upper Limit = x̅ + E =    123.93 -   4.924   =   128.857
confidence interval is (   119.01 < µ <   128.86 )  

d)

larger the confidence level , larger the critical value and hence, larger the cnfidence interval

so, 95%  confidence interval be wider than the interval you found in Part (c)

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   5          
't value='   tα/2=   2.5706   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   5.985   / √   6   =   2.4434
margin of error , E=t*SE =   2.5706   *   2.4434   =   6.281
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    123.93   -   6.281   =   117.652
Interval Upper Limit = x̅ + E =    123.93   -   6.281   =   130.214
confidence interval is (   117.65 < µ <   130.21 )