Answer:
Pvalue > α ; We fail to reject the Null; hence no significant difference exists in the average prices of the two supermarkets.
(−68.1095 ; 57.8545)
Step-by-step explanation:
Given the data:
Week Advertiser ($) Competitor ($) 1 254.27 255.93 2 240.72 255.65 3 231.91 255.12 4 234.15 261.28
Difference, d :
254.27 - 255.93 = - 1 66
240.72 - 255.65 = - 14.93
231.91 - 255.12 = - 23.21
234.15 - 261.28 = - 27.13
d = -1.66, -14.93, 23.21, -27.13
Using calculator :
Mean difference , dbar = - 5.1275
Standard deviation of mean , Sd = 21.566
H0 : μd = 0
H1 : μd ≠ 0
Test statistic :
(dbar - μd) / s/√n
(-5.1275 - 0) / (21.566/√4)
-5.1275 / 10.783
= −0.475517
Using the Pvalue from Tscore Calculator ; df = 4 - 1 = 3
Pvalue = 0.66659
Pvalue > α ; We fail to reject the Null; hence no significant difference exists in the average prices of the two supermarkets.
Confidence interval :
dbar ± standard Error
Standard Error = Tcritical * Sd/√n
dbar = - 5.1275
Tcritical at 0.01 = 5.941
Standard Error = 5.8409 * 21.566/√4 = 62.982
Confidence interval :
Lower boundary : - 5.1275 - 62.982 = −68.1095
Upper boundary = - 5.1275 + 62.982 = 57.8545
(−68.1095 ; 57.8545)
The true population on difference exist within the interval
