Question

A survey was conducted by an independent price-checking company to check an advertiser's claim that it had lower prices than its competitors. The average weekly total, based on the prices of approximately 95 items, is given for this chain and for its competitor recorded during four consecutive weeks in a particular month.
Week Advertiser ($) Competitor ($) 1 254.27 255.93 2 240.72 255.65 3 231.91 255.12 4 234.15 261.28
a. Is there a significant difference in the average prices for these two difference supermarket chains?b. What is the approximate p-value for the test conducted in part a?c. Construct a 99% confidence interval for the difference in the average prices for the two supermarket chains. Interpret this interval.

Answer 1

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