Answer:
The P-value is 0.04542 which is less than 0.05
Since the calculated value of t does not fall in the critical region so we accept H0 that there is no price difference between the two stores.
Explanation:
Store 1 Store 2 Difference (Store 2-1) d²
Product A 5.92 5.85 -0.07 0.0049
Product B 7.53 7.93 0.4 0.16
Product C 3.75 3.98 0.23 0.0529
Product D 1.81 1.71 -0.1 0.01
Product E 1.71 1.96 0.25 0.0625
Product F 2.86 2.58 - 0.28 0 .0784
Product G 4.79 4.74 -0.05 0.0025
Product H 3.18 3.69 0.51 0.2601
Product I 3.02 2.87 - 0.15 0.0225
Product J 3.77 3.68 -0.09 0.0081
∑ 0.65 0.6619
1. The degrees of freedom = n-1= 10-1= 9
2. The significance level is 0.05
3.The test statistic is
t= d`/sd/√n
4. The critical region is ║t║≥ t (0.025,9) = 2.262
The null and alternate hypotheses is
H0 : ud= 0 against Ha: ud≠ 0 (two tailed test)
5. Calculations
d`= ∑di/n= 0.65/10= 0.065
Sd²= ∑(di-d`)²/n-1 = 1/n-1 [∑di²- (∑di)²n]
Sd²= 1/9[0.6619-(0.65)²/10] = [0.6619-0.4225/9]= 0.0266
Sd= 0.163
Therefore
t= d`/ sd/√n
t= 0.065/ 0.163/√10
t= 0.3987/3.1622=0.1261
6. Conclusion
Since the calculated value of t does not fall in the critical region so we accept H0 that there is no price difference between the two stores.
Now the P- value:
The P-value is the rejection region. In this test it is on both sides and would add up to be (2.262 + 2.262)/100= 4.542/100
The P-value is 0.04542 which is less than 0.05