Suppose 8 cups of coffee are served to a subject, where for each cup a coin is f

Question

Suppose 8 cups of coffee are served to a subject, where for each cup a coin is flipped to determine if regular or decaf coffee is poured. Also suppose that the subject correctly distinguishes 7 of the 8 cups. Test the hypothesis that the subject has no sensory discrimination. What is your p-value? Suppose instead that the cups were presented to the subject in pairs, where each pair had one cup of decaf and one regular, and the order was determined by the flip of a coin. How many cups of coffee would you need to serve the subject to obtain a p-value for one mistake that is roughly the same size as in the previous design?

Explanation / Answer

Let us denote p = Proportion of cups distinguished correctly = 7/8 = 0.875

If subject has NO sensory discrimination, it means we will test the hypothesis that

H0 : P=0.5 Vs H1: P > 0.5

Test statistic under H0 is

Z = (p-P)/SE(p) ~ N(0,1) where SE(p) = P(1-P)/n = 0.5*0.5/8 = 0.176777

Z = (0.875-0.5)/0.176777 = 2.12132

P value = P[Z > 2.12132] = 0.016947

Conclusion: At 1% LOS we can say that subject do not have sensory discrimination.

Now if cups are presented in pair with each type of one cup, then there will be only 4 pairs. Probability of discriminating each pair correctly is 0.5.

Define r.v. X as Number of pairs submitted for discrimination.

Here X: 0,1,2,….,4

X ~ B(n, p=0.5)

Aim is to find ‘k’ such that p-value is same as above.

Here assume that ‘n’ is large so that XàN(Mean = np, Variance = npq)

i.e. X àN(2, 1)

Let us construct the probabilistic equation as

P[X>k]= 0.016947

P[Z > (k-2)] = 0.016947

From Normal Table we have Z table= 2.12132

Thus (k-2) = 2.12132 it implies that k = 4.12132 = 4

It means getting p-value same as above is NOT possible.

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