To test whether a coin is a fair coin, we decide to flip this coin 10 times and

Question

To test whether a coin is a fair coin, we decide to flip this coin 10 times and count the numbers of Heads (H) and Tails (T) we observe. Let X be the number of Hs observed from 50 tosses and p be the actual probability of getting H from flipping this coin. If this is a fair coin, then p = .5. Suppose the significance level is .05.

Here are the hypotheses:

H0: p=.5

Ha: p.5

1-a: Compute the p-value if we have observed 40 Ts and draw your conclusion whether this is a fair coin.

1-b: Compute the probability of Type II error given p = 0.55 for n = 50, 75, and 100, respectively.

1-c: What impact does n have on probability of Type II error?

Explanation / Answer

H0: p = 0.5

Ha: p 0.5

(a) Here number of Ts = 40

Hs = 10

so p -value = 2 * Pr(Hs < = 10) = 2 * BIN (H <= 10; 50; 0.5) = 2.386 x 10-5

multiplied by 2 because two tailed test.

Here that probability is less than 0.05 so we can say that the coin is not fair

(1-b) Here type II error is not rejecting null hypothesis even if it is true.

for n = 50.

proportion below which we shall not reject the null hypotheis is = p0 +Z95% sqrt [p0 (1-p0)/N]  

= 0.5 + 1.96 * sqrt [0.5 *0.5/50] = 0.6386

Pr(Type II error ) = Pr(p^ < 0.6386) where true population proportion p = 0.55 and

standard error of proportion = sqrt(0.55 *0.45/50) = 0.0704

Pr(Type II error ) = Pr(p^ < 0.6386) = Pr(p^ < 0.6386; 0.55; 0.0704)

Z = (0.6386 - 0.55)/ 0.0704 = 1.26

Pr(Type II error ) = Pr(Z < 1.26) = 0.9680

for n = 75

proportion below which we shall not reject the null hypotheis is = p0 +Z95% sqrt [p0 (1-p0)/N]  

= 0.5 + 1.96 * sqrt [0.5 *0.5/75] = 0.6132

Pr(Type II error ) = Pr(p^ < 0.6132 where true population proportion p = 0.55 and

standard error of proportion = sqrt(0.55 *0.45/75) = 0.0574

Pr(Type II error ) = Pr(p^ < 0.6132) = Pr(p^ < 0.6132; 0.55; 0.0574)

Z = (0.6132 - 0.55)/ 0.0574= 1.10

Pr(Type II error ) = Pr(Z < 1.10) = 0.8646

for N = 100

proportion below which we shall not reject the null hypotheis is = p0 +Z95% sqrt [p0 (1-p0)/N]  

= 0.5 + 1.96 * sqrt [0.5 *0.5/100] = 0.598

Pr(Type II error ) = Pr(p^ < 0.598) where true population proportion p = 0.55 and

standard error of proportion = sqrt(0.55 *0.45/100) = 0.04975

Pr(Type II error ) = Pr(p^ < 0.598) = Pr(p^ < 0.598; 0.55; 0.04975)

Z = (0.598 - 0.55)/ 0.04975= 0.9648

Pr(Type II error ) = Pr(Z < 0.9648) = 0.8327

1-c As sample size increases the probability of type II error decreases.

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