Question

Ronald Fisher, widely considered the father of modern statistics, famously offered a cup of tea to fellow scientist Muriel Bristol. Bristol declined taking the tea saying that she preferred the flavor when the milk was poured into the cup before the tea. Fisher did not believe that the order of pouring (milk first or tea first) could affect the flavor. Bristol claimed she could tell the difference. A test was then devised to determine whether Bristol truly could distinguish between tea poured before or after the milk. Fisher reported the experiment in his 1935 book The Design of Experiments.

To test Bristol’s claim, 8 cups of tea were prepared. Four cups were poured milk first and the other four were poured tea first. These were then randomly arranged and offered to Bristol. At each cup, she will taste the tea and then identify whether the milk was poured first or not. The outcome of each classification is either correct or incorrect. Since the experiment seeks to test the claim that she is not simply guessing, the assumption must be that Bristol does not actually have the ability to distinguish between the different cups.

1) Under the assumption that Bristol cannot distinguish between the cups, what is the probability that she correctly classifies a single cup?

2)What is the mean and standard deviation of the distribution?

3) Use the range rule of thumb to determine if the following numbers of correct guesses would be considered significantly high, significantly low or insignificant. a) 5 correct guesses b) 2 correct guesses c) 8 correct guesses

Answer 1

Solution

Back-up Theory

If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where n = number of trials and

p = probability of one success, then, probability mass function (pmf) of X is given by

p(x) = P(X = x) = (ⁿC_x)(p^x)(1 - p)^{n – x}, ………….........................................................................………..(1)

[This probability can also be directly obtained using Excel Function: Statistical, BINOMDIST]....................………….(1a)

Mean (average) of X = E(X) = µ = np….....................................................................……………………………………………..(2)

Variance of X = V(X) = σ² = np(1 – p)………….................................................................………………………………………..(3)

Standard Deviation of X = SD(X) = σ = √{np(1 – p)} ……......................................................…………………………………...(4)

Range Rule of Thumb: Range approximately equal to four times the standard deviation.... (4)

Now to work out the solution,

Part (a)

Under the assumption that Bristol cannot distinguish between the cups, she would have randomly guessed and in that case,

the probability that she correctly classifies a single cup is ½. Answer 1

Part (b)

If X = number of cups Bristol guessed correctly, under the assumption that she cannot distinguish between the cups, X ~ B(8, 0.5). Hence, vide (2) and (4),

Mean = 8 x 0.5 = 4 Answer 2

Standard deviation = √(8 x 0.5 x 0.5) = 1.414 Answer 3

Part (c)

Vide (4), the expected range for X must be 4 x 1.414 = 5.6. This implies 2.8 on either side of the mean, i.e., 1.2 to 6.8 would be treated as insignificant, more than 6.8 as significantly high and less than 1.2 as significantly low. Thus,

(a) and (b) are insignificant and (c) is significantly high. Answer 4

DONE