Question

The author drilled a hole in a die and filled it with a lead​ weight, then proceeded to roll it 199 times. Here are the observed frequencies for the outcomes of​ 1,2,3,4,5, and 6​ respectively: 28​, 29​, 47​, 40​, 22​, 33. Use a 0.05 significance level to test the claim that the outcomes are not equally likely. Does it appear that the loaded die behaves differently than a fair​ die?

Answer 1

Answer with explanation:

An Unbiased Dice is Rolled 199 times.

Frequency of outcomes 1,2,3,4,5,6 are=28​, 29​, 47​, 40​, 22​, 33.

Probability of an Event

`=\frac{\text{Total favorable Outcome}}{\text{Total Possible Outcome}}\\\\P(1)=(28)/(199)\\\\P(2)=(29)/(199)\\\\P(3)=(47)/(199)\\\\P(4)=(40)/(199)\\\\P(5)=(22)/(199)\\\\P(6)=(33)/(199)\\\\\text{Dice is fair}\\\\P(1,2,3,4,5,6}=(33)/(199)`

→→→To check whether the result are significant or not , we will calculate standard error(e) and then z value

1.

`(e_(1))^2=(P_(1))^2+(P'_(1))^2\\\\(e_(1))^2=[(28)/(199)]^2+[(33)/(199)]^2\\\\(e_(1))^2=(1873)/(39601)\\\\(e_(1))^2=0.0472967\\\\e_(1)=0.217478\\\\z_(1)=(P'_(1)-P_(1))/(e_(1))\\\\z_(1)=((33)/(199)-(28)/(199))/(0.217478)\\\\z_(1)=(5)/(43.27)\\\\z_(1)=0.12`

→→If the value of z is between 2 and 3 , then the result will be significant at 5% level of Significance.Here value of z is very less, so the result is not significant.

2.

`(e_(2))^2=(P_(2))^2+(P'_(2))^2\\\\(e_(2))^2=[(29)/(199)]^2+[(33)/(199)]^2\\\\(e_(2))^2=(1930)/(39601)\\\\(e_(2))^2=0.04873\\\\e_(2)=0.2207\\\\z_(2)=(P'_(2)-P_(2))/(e_(2))\\\\z_(2)=((33)/(199)-(29)/(199))/(0.2207)\\\\z_(2)=(4)/(43.9193)\\\\z_(2)=0.0911`

Result is not significant.

3.

`(e_(3))^2=(P_(3))^2+(P'_(3))^2\\\\(e_(3))^2=[(47)/(199)]^2+[(33)/(199)]^2\\\\(e_(3))^2=(3298)/(39601)\\\\(e_(3))^2=0.08328\\\\e_(3)=0.2885\\\\z_(3)=(P_(3)-P'_(3))/(e_(3))\\\\z_(3)=((47)/(199)-(33)/(199))/(0.2885)\\\\z_(3)=(14)/(57.4279)\\\\z_(3)=0.24378`

Result is not significant.

4.

`(e_(4))^2=(P_(4))^2+(P'_(4))^2\\\\(e_(4))^2=[(40)/(199)]^2+[(33)/(199)]^2\\\\(e_(4))^2=(3298)/(39601)\\\\(e_(4))^2=0.06790\\\\e_(4)=0.2605\\\\z_(4)=(P_(4)-P'_(4))/(e_(4))\\\\z_(4)=((40)/(199)-(33)/(199))/(0.2605)\\\\z_(4)=(7)/(51.8555)\\\\z_(4)=0.1349`

Result is not significant.

5.

`(e_(5))^2=(P_(5))^2+(P'_(5))^2\\\\(e_(5))^2=[(22)/(199)]^2+[(33)/(199)]^2\\\\(e_(5))^2=(1573)/(39601)\\\\(e_(5))^2=0.03972\\\\e_(5)=0.1993\\\\z_(5)=(P'_(5)-P_(5))/(e_(5))\\\\z_(5)=((33)/(199)-(22)/(199))/(0.1993)\\\\z_(5)=(11)/(39.6610)\\\\z_(5)=0.2773`

Result is not significant.

6.

`(e_(6))^2=(P_(6))^2+(P'_(6))^2\\\\(e_(6))^2=[(33)/(199)]^2+[(33)/(199)]^2\\\\(e_(6))^2=(2178)/(39601)\\\\(e_(6))^2=0.05499\\\\e_(6)=0.2345\\\\z_(6)=(P'_(6)-P_(6))/(e_(6))\\\\z_(6)=((33)/(199)-(33)/(199))/(0.2345)\\\\z_(6)=(0)/(46.6655)\\\\z_(6)=0`

Result is not significant.

⇒If you will calculate the mean of all six z values, you will obtain that, z value is less than 2.So, we can say that ,outcomes are not equally likely at a 0.05 significance level.

⇒⇒Yes , as Probability of most of the numbers that is, 1,2,3,4,5,6 are different, for a loaded die , it should be equal to approximately equal to 33 for each of the numbers from 1 to 6.So, we can say with certainty that loaded die behaves differently than a fair​ die.