In: Statistics and Probability
A consumer group wished to compare the price of food items at two supermarkets A and B. A random sample of 8 items was selected and the cost of the items at each supermarket was recorded on a particular day.
Item 1 2 3 4 5 6 7 8
A $3.60 $3.73 $3.48 $2.43 $2.81 $2.66 $3.62 $2.40
B $3.38 $3.77 $3.50 $2.55 $2.82 $2.65 $3.74 $2.31
(a) test at α = .05 if "on average" the prices offered at the two supermarkets differ. (d= .0013, sd = .1118)
(b) Conduct the test in (a) using appropriate confidence interval
Number | A | B | Difference | |
3.6 | 3.38 | 0.22 | 0.04785156 | |
3.73 | 3.77 | -0.04 | 0.00170156 | |
3.48 | 3.5 | -0.02 | 0.00045156 | |
2.43 | 2.55 | -0.12 | 0.01470156 | |
2.81 | 2.82 | -0.01 | 0.00012656 | |
2.66 | 2.65 | 0.01 | 7.6563E-05 | |
3.62 | 3.74 | -0.12 | 0.01470156 | |
2.4 | 2.31 | 0.09 | 0.00787656 | |
Total | 24.73 | 24.72 | 0.01 | 0.0874875 |
To Test :-
H0 :-
H1 :-
Test Statistic :-
t = 0.0316
Test Criteria :-
Reject null hypothesis if
Result :- Fail to reject null hypothesis
Confidence Interval :-
Lower Limit =
Lower Limit = -0.0924
Upper Limit =
Upper Limit = 0.0949
95% Confidence interval is ( -0.0924 , 0.0949 )
Since 0 lies in the interval, we can conclude to accept null hypothesis
Conclusion :- There is no sufficient evidence to conclude that the average prices offered at the two supermarkets differ.