1. A gambler wants to test whether the die she has is fair so she rolls it 300 t

Question

1. A gambler wants to test whether the die she has is fair so she rolls it 300 times and records the outcomes as Rol 1 2345 6 Count 48 34 53 47 58 60 meaning she rolled a 1 on the die 48 times, etc (a) Give a 95% confidence interval for the probability of the events: i. The die rolls a 1 ii. The die rolls a 2. iii. The die rolls either a 1 or a 2 (b) Using a binomial model, calculate the P-value for testing whether rolling a 1, 2, or 3 is significantly less likely than a 4, 5, or 6 (c) Using a 2 goodness of fit test, test whether the 6 probabilities are all the same. Use an 0.05 level test.)

Explanation / Answer

(a) For die rolls a 1

p^ = 48/300 = 0.16

Here population proportion is known so,

stanard error of the proportion = sqrt [1/6 * 5/6 / 300] = 0.0215

95% confidence interval = p^ +- Z95% sqrt[p^ * (1 -p^)/ N]

= 0.16 +- 1.96 * 0.0215

= (0.1178, 0.2022)

(ii) For die rolls a 2

p^ = 34/300 = 0.1133

Here population proportion is known so,

stanard error of the proportion = sqrt [1/6 * 5/6 / 300] = 0.0215

95% confidence interval = p^ +- Z95% sqrt[p^ * (1 -p^)/ N]

= 0.1133 +- 1.96 * 0.0215

= (0.07, 0.155)

(c) For die rolls either 1 or 2

p^ = 82/300 = 0.2733

Here population proportion is known so,

stanard error of the proportion = sqrt [2/6 * 4/6 / 300] = 0.0272

95% confidence interval = p^ +- Z95% sqrt[p^ * (1 -p^)/ N]

= 0.2733 +- 1.96 * 0.0272

= (0.22, 0.3266)

(b) H0 : p1,2,3 = p4,5,6  = 0.5 = p0

Ha : p1,2,3 < p4,5,6

Standard error of the proportion se= sqrt [p0 * (1 - p0 )/N] = sqrt [0.5 * 0.5/ 300] = 0.0289

Here p^1,2,3 = (48 + 34 + 53)/ 300 = 0.45

so test statistic

Z = (p^ - p0)/ se = (0.45 - 0.50)/ 0.0289 = - 1.73

P - value = Pr (Z < -1.73) = 0.0418 < 0.05

so, here we can deduce that null hypothesis is incorrect and we shall say that rolling 1,2,3 is significantly less likely than a 4,5,or 6.

(c) The Observed and Expected Table

Here Expected value for each roll = 300 * (1/6) = 50

so, X2 = 8.84

so dF = 6-1 = 5 and alpha = 0.05

X2cr = 11.07

p - value = CHITEST (Expecte, Observed) = 0.1156 > 0.05

so we cannot reject the null hypothesis and can conclude that all 6 probabibilities are same.

Roll Observed Count (O) Expected Count (E) (O-E)^2/E 1 48 50 0.08 2 34 50 5.12 3 53 50 0.18 4 47 50 0.18 5 58 50 1.28 6 60 50 2 sum 8.84

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