1.
Discount stores often introduce new merchandise at a special low price toinduce people to try it. But in the mid-1960s a prominent psychologistpredicted that in the long run this practice would actually reduce sales.With the cooperation of a discount chain (I think it was K-Mart), an experimentwas performed in 1968 to test this theory. A representative sample of 120stores was chosen, and the stores were arranged into 60 pairs, matchedaccording to characteristics like sales volume and location. These stores didnot advertise, and displayed their merchandise in similar ways. A new kind ofcookie was introduced in all 120 stores. Within each pair of stores, one waschosen at random to introduce the cookies at the special low price of 49 centsa box, with the price increasing to 69 cents after two weeks; the other store inthe pair introduced the cookies at the regular price of 69 cents a box. Totalsales (in cases) of the cookies were computed for each store for six weeksfrom the time they were introduced; the results are given below.
pair discount standard difference
number sales sales (discount - standard)
======
1 851 916 -65
2 903 1004 -101
. . . .
. . . .
60 787 699 +88
======
mean 854 923 -69
SD 58 157 150
Does this evidence support or refute the psychologist's theory?What was the point of pairing the stores in the way they did?Explain briefly
n = n1 = n2 60
d-bar = -69
s (of d) = 150
Hypotheses:
If the psychologist is correct, then the discount sales must be less than the standard sales. So, the mean difference
d-bar = (discount sales - standard sales)< 0. This will be the alternative hypothesis. The null hypothesis will be that
d-bar ≥ 0.
Ho: d-bar ≥ 0
Ha: d-bar < 0
Decision Rule:
α = 0.05
Degrees of freedom = 60 - 1 = 59
Critical t- score = -1.671093033
Reject Ho if t < -1.671093033
Test Statistic:
SE = s/n = 150/√60 = 19.36491673
t = d-bar/SE = -69/19.3649167310371 = -3.563144679
p- value = 0.000733058
Decision (in terms of the hypotheses):
Since -3.563144679 < -1.671 we reject Ho and accept Ha
Conclusion (in terms of the problem):
It appears that discount sales are significantly less than standard sales. The findings support the psychologist’s theory.
Notes:
(a) Since the sample size is large, the data for the sales as well as the difference in sales can be expected to follow normal distribution. Sampling is also random. Thus, requirements for a paired t- test are met.
(b) The pairing of the stores according to sales volume, location etc introduces some sense of similarity and permits us tobetter compare things. It also eliminates potential confounding factors.
2.
In 1969, the well-known pediatrician Dr. Benjamin Spock came to trial beforea judge named Ford in Boston's Federal court house. He was charged withconspiracy to violate the Military Service Act (in addition to his work on childdevelopment he was active in anti-war protests in the 60s). A lawyer writingabout the case that same year in the Chicago Law Review said about thecase, "Of all defendants at such trials, Dr. Spock, who had given wise andwelcome advice on child-bearing to millions of mothers, would have likedwomen on his jury."The jury was drawn from a panel of 350 persons, called a venire, selected byJudge Ford's clerk. This venire included only 102 women, even though 53%of the eligible jurors in the district were female. At the next stage in selectingthe jury to hear the case, Judge Ford chose 100 potential jurors out of these350 people. His choices included only 9 women.If 350 people are chosen from all the eligible jurors in the district, how likely isit that the sample will include 102 women or fewer?If 100 people are chosen at random without replacement from a group ofpeople consisting of 102 women and 248 men, what is the chance that thesample will include 9 women or fewer? (Hint: remember the correction factor,if relevant.)What do you conclude about the impartiality of Judge Ford's selectionprocess?Explain briefly.
n = 350
p = 102/350 = 0.2914
SE of proportion = √[π * (1 - π)/n] = √[(0.53 * 0.47)/350] = 0.0267
z = (p - π)/SE = (0.2914 - 0.53)/0.0267 = -8.94
P(p ≤ 0.2914) = P(z < -8.94) = 0
The probability that 102 or fewer women would be selected in a venire of 350 persons is 0. This means it is highly unlikely that 102 or fewer women would be selected from a venire of 350 persons. (There ought to have been more than 102 women selected out of 350.)
(b) π = 102/350 = 0.2914
n = 100
p = 9/100 = 0.09
Population correction factor = √[(N - n)/(N - 1)] = √[(350 - 100)/(350 - 1)] = 0.846
SE of proportion = 0.846 * √[π * (1 - π)/n] = √[(0.2914 * 0.7086)/100] = 0.038
z = (p - π)/SE = (0.09 - 0.2914)/0.038 = -5.3
P(p ≤ 0.09) = P(z < -5.3) = 0
The probability that the sample will include 9 women or fewer is 0. This means it is highly unlikely that 9 or fewer women would ultimately be on the jury. (There ought to have been more than 9 women on the jury.)
Note:
The results in (a) and (b) above clearly show that Judge Ford’s selection was biased against women in both the stages. He was not selecting randomly. There is evidence that he was discriminating against women in the selection process.