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 Roll 1 23 4 5 6 Count 48 34 53 47 58 60 meaning she rolled a 1 on the die 48 times, ete. (a) Give a 95% confidence interval for the probability of the events: i. The die rolls a1 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 st test whether the 6 probabilitie, are all the same (Use an a-aas level test.)

Explanation / Answer

PART A.
i.
given that,
possibile chances (x)=48
sample size(n)=300
success rate ( p )= x/n = 0.16
CI = confidence interval
confidence interval = [ 0.16 ± 1.96 * Sqrt ( (0.16*0.84) /300) ) ]
= [0.16 - 1.96 * Sqrt ( (0.16*0.84) /300) , 0.16 + 1.96 * Sqrt ( (0.16*0.84) /300) ]
= [0.119 , 0.201]
ii.
possibile chances (x)=34
sample size(n)=300
success rate ( p )= x/n = 0.113
CI = confidence interval
confidence interval = [ 0.113 ± 1.96 * Sqrt ( (0.113*0.887) /300) ) ]
= [0.113 - 1.96 * Sqrt ( (0.113*0.887) /300) , 0.113 + 1.96 * Sqrt ( (0.113*0.887) /300) ]
= [0.077 , 0.149]
iii.
given that,
possibile chances (x)=48+34=82
sample size(n)=300
success rate ( p )= x/n = 0.273
CI = confidence interval
confidence interval = [ 0.273 ± 1.96 * Sqrt ( (0.273*0.727) /300) ) ]
= [0.273 - 1.96 * Sqrt ( (0.273*0.727) /300) , 0.273 + 1.96 * Sqrt ( (0.273*0.727) /300) ]
= [0.223 , 0.324]

PART C.

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calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above

Expected frequency = 300/6 = 50

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set up null vs alternative as

null, Ho: no relation b/w X and Y OR X and Y are independent

alternative, H1: exists a relation b/w X and Y OR X and Y are dependent And all are similar

level of significance, = 0.05

from standard normal table, chi square value at right tailed, ^2 /2 =11.07

since our test is right tailed,reject Ho when ^2 o > 11.07

we use test statistic ^2 o = (Oi-Ei)^2/Ei

from the table , ^2 o = 8.84

critical value

the value of |^2 | at los 0.05 with d.f n-1 = 5 is 11.07

we got | ^2| =8.84 & | ^2 | =11.07

make decision

hence value of | ^2 o | < | ^2 | and here we reject Ho

ANSWERS

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null, Ho: no relation b/w X and Y OR X and Y are independent

alternative, H1: exists a relation b/w X and Y OR X and Y are dependent

test statistic: 8.84

critical value: 11.07

decision: reject Ho

we conlcude that probabilities are dependent and could be equal

Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 48 50 -2 4 0.08 34 50 -16 256 5.12 53 50 3 9 0.18 47 50 -3 9 0.18 58 50 8 64 1.28 60 50 10 100 2 8.84

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