Question

1. You are observing a series of independent matches between two teams. After 500 matches, Team A has won 300 matches and Team B has won the remaining 200 matches (there are no ties) Let P(A) be the probability that Team A wins a given match [5] (a) Find an unbiased estimate of P(A [51 (b) Find an interval [a, b such that P(A) E [a, b] with probability 0.95 5] (c) Suppose that the interval in part (b) is too large. How many more matches would you have to observe to cut the length of that interval in half? bility that Team B wins more than 220 of the next 500 matches.

Answer 1

Solution

Back-up Theory

Probability can be estimated by relative frequency or proportion of occurrences. ……………………. (1)

Interpretation of Confidence Interval

[a, b] is 100(1 - α) % Confidence Interval for a population parameter, say θ, implies that

Probability θ ∈ [a, b] = (1 - α) ………………………………………………………………………….. (2)

If X ~ B(n, p) np ≥ 10 and np(1 - p) ≥ 10, by application of Central Limit Theorem,

[(X – np)/√{np(1 - p)}] ~ N(0, 1) [approximately] …………………………………………………… (2a)

Binomial probability of

Now, to work out the solution,

Part (a)

Vide (1), unbiased estimate of P(A) = proportion of matches won by A

= 300/500

= 0.6 Answer 1

Part (b)

Vide (2), given condition => [a, b] is the 95% Confidence Interval for P(A), i.e., the population proportion.

Now, 100(1 - α) % Confidence Interval for the population proportion, p is:

p_cap ± MoE, …………………………………………………………………………………………….. (3)

where

MoE = Z_α/2[√{p_cap(1 –p_cap)/n}] ………………………………………………………………………...(4)

with

Z_α/2 is the upper (α/2)% point of N(0, 1),

pcap = sample proportion, and

n = sample size.

So, 95% Confidence Interval for P(A) = 0.6 ± MoE,

where

MoE = 1.96√{0.6 x 0.4/500} [1.96 = upper 2.5% point of N(0, 1), obtained using Excel Function: Statistical

                                                 NORMSINV ]

= 1.96 x 0.0219

= 0.0429.

Hence, [a, b] = [0.5571, 0.6429] Answer 2

Part (c)

Vide (3), length of the confidence interval is solely determined by the MoE, which in this case is 0.0429 [refer the working of Part (b)].

If this has to be halved, we should have new MoE = 0.02145.

Again vide (4), we should have: 1.96√{0.6 x 0.4/n} = 0.02145, where n is the sample size which we want to determine.

So, n = (1.96/0.02145)² x 0.24

= 2004

Thus, 1504 more matches are to be observed. Answer 3

Part (d)

From Part (a), P(A) = 0.6. Since there are no ties, P(B) = 1 – P(A) = 0.4.

If X = number of matches B wins in 500 matches, then X ~ B(500, 0.4).

So, the required probability

= P(X > 220)

= P[Z > {(220 – 200)/√(200 x 0.6)]

= P(Z > 1.8257)

= 0.034 [obtained using Excel Function: Statistical NORMSDIST ] Answer 4

DONE