Question

‘Lady tasting tea’ is a randomized experiment devised by Ronald Fisher and reported in his book The Design of Experiments (1935). The lady in question (Muriel Bristol) claimed that she can tell whether the tea or the milk was added first to a cup. Fisher proposed to give her eight cups, four of each variety, in random order. One could then ask what the probability was for her getting the specific number of cups she identified correct, but just by chance. In this problem we are going to modify and simulate ‘The Lady tasting tea’ experiment. Suppose that you are going to randomly selected a student and conduct the similar experiment; 10 cups (instead of 8) of tea of which some have milk added first and others have milk added last. We are going to assume that this student, unlike Ms Bristol, does not have the skill to tell if the milk was added first or last.

1. (Lady tasting tea Problem) What is the probability that the student correctly guesses if a given cup of tea has milk added first or last?

2. (Lady tasting tea Problem) For this student, what is the expected number of correct guesses out of 10? What about incorrect guesses?

3. (Lady tasting tea Problem) Would you be surprised if the student guesses 6 out of 10 cups correctly (60% correct rate)? Would you conclude that this student has the skill to tell if the milk was added first or last? Or would you conclude that he/she was just “lucky”?

4. (Lady tasting tea Problem) Suppose that you prepare 100 cups of tea instead of 10. Would you be surprised if the student guesses 60 out of 100 cups correctly (still 60% correction rate)? Would you conclude that this student has the skill to tell if the milk was added first or last? Or would you conclude that he/she was just “lucky”?

Answer 1

This is a question related to simple visualization of the activities conducted and the corresponding probabilities calculated.

The student selected is a random act and there is NO biasness in tea tasting abilities of the student selected.

1. ). Since the student does not have any special prowess to select a cup of milk with milk first ,we can consider the student to be an unbiased sample. Thus he will have an equal chance of selection or not selecting a cup of tea having milk added first.

In other words, the probability in this case will be same for both selection or rejection.

or, P(selction)+P(wrong selection)=1 and

P(selction)=P(wrong selection)

Thus we have

P(selection)=1/2 =0.50

Thus the probability that a given student with no special abilities ot guess correctly the milk first in tea , in a given cup of tea is 0.50.

2). In a given chance on an avarage, the student has a 50% chance of making correct guess.

Assuming this to be valid for 10 trials, (10 cup of tea),

The expected avarage number of guess that he will call correctly is P(selection)*10

=1/2*10

=5

Thus on an average,the unbiased student can select 5 cups correctly for milk first.

Hence total number of cups guessed in correctly out of 10 given cups = Total number of cups -total cups guessed correctly on average

=10-5

=5 cups for incorrect guesses

3). Now we are asked ot check if the act of the student selecting 6 cups out of 10 correctly is significant or not.

For this we shall conduct a hypothesis test wherein we assume that the given selection of 6/10 cups is not special /significant

( I simulated the hypothesis testing on an online simulator and got the following results)

Hence we can say that the student selecting 6 /10 correctly is not significant and the guy has simply got lucky.  

4). Now in this case, N =100 and n(number of favourable caes)=60

We will again run the hypotheiss testing as follows and get

Hence we can say that the student selecting 60 /100 correctly is significant and the guy has got special prowess to do so.