2. We want to compare the variances in two dealerships sales of our A and B dealership. Use the following table to develop a comparison about the two population variances. Prepare the information for Hypothesis Testing.
Dealership A
Team Member | Average dollar sale | xbar | x-xbar | x-xbar^2 |
---|---|---|---|---|
1 | 25000 | |||
2 | 25500 | |||
3 | 23750 | |||
4 | 25250 | |||
5 | 24250 | |||
6 | 24750 | |||
7 | 25750 | |||
8 | 24500 | |||
9 | 25375 | |||
10 | 24625 | |||
Dealership B
Team member | Average dollar sale | xbar | x-xbar | x-xbar^2 |
1 | 24625.00 | |||
2 | 24735.00 | |||
3 | 23631.25 | |||
4 | 24745.00 | |||
5 | 24189.38 | |||
6 | 24675.75 | |||
7 | 24720.00 | |||
8 | 24426.50 | |||
9 | 24378.75 | |||
10 | 24378.85 |
244486.63
1. Mean of first set?
2. Mean of second set?
3. Variance of first set?
4. Variance of second set?
5. Observations of the first set?
6. Observations of the second set?
7. Degrees of freedom first set?
8. Degrees of freedom second set?
9. Calculated test statistic? (4 decimal places)
10. P=Value? (4 decimal places)
11. Critical value? (4 decimal places)
1. Mean of first set = 24875
2. Mean of second set = 24450.55
3. Variance of first set = 381944.44
4. Variance of second set = 119077.84
5. Observations of the first set = 10
6. Observations of the second set = 10
7. Degrees of freedom first set = 9
8. Degrees of freedom second set = 9
9. Calculated test statistic = F = 3.2075
10. P-Value = 0.0488
11. Critical value = 3.1789
Since F> Critical value and also P-value is less than
0.05, we reject H0. We conclude that there is enough evidence to
suggest that the two population variances of dealership sales are
different here.
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