Question

1. -18 points My Notes A six-sided die is rolled 120 times. We assume these are a simple random sample of its rolls. Fill in the expected frequency column. Then, conduct a hypothesis test at the 5% level to determine if the die is fair. The data below are the resultof the 120 rolls Face Value Frequency Expected Frequency 14 34 16 15 25 16 4 6 Enter an exact number

a.) Are the conditions for running this test stratified?

- Yes, all observed counts are bigger than 5.

- Yes, the mean number of counts is bigger than 10.

- Yes, the total number of rolls is bigger than 10.

- Yes, all expeced counts are bigger than 10.

b.) State the null hypothesis.

- The data fit the distribution for a fair six-sided die.

- The data do not fit the distribution for fair six-sided die.

c.) What are the degrees of freedon?

d.) State the distribution to use for the test.

- X 2/5

- X²₆

- t₆

- t₅

e.) What is the test statistic? (Round your answer to two decimal places.)

f.) What is the p-value? (Round your answer to four decimal places.)

Explain what the p-value means for this problem.

- if H_O is false, then there is a chance equal to the p-value that the value of the test statistic will be equal to or less than the calculated value.

- if H_O is true, then there is a chance equal to p-value that the volume of the test statistic will be equal to or greater than the calculated value.

- if H_O is false, then there is a chance equal to p-value that the value of the test statistic will be equal to or greater than the calculated value.

- if H_O is true, then there is a chance equal to p-value that the value of the test statistic will be equal to or less than the calculated value.

g.) Indicate the correct decison ("reject" or "do not reject" the null hypothesis) and write the appropriate conclusion.

    i) Alpha:    a = ________

   ii) Decision:

          - reject the null hypothesis

         - do not reject the null hypothesis

    iii) Reason for decision:

            - Since a > p-value, we reject the null hypothesis.

           - Since a < p-value, we do not reject the null hypothesis.

          - Since a > p-value, we do not reject the null hypothesis.

          - Since a < p-value, we reject the null hypothesis.

     iv) Conclusion:

          - There is sufficient evidence to conclude that the data do not fit the distribution for a fair six-sided die.

          - There is not sufficient evidene to conclude that the data do not fir the distribution for a fair six-sided die.

Answer 1

Here we have given that six sided die is rolled 120 times.

For fair die the probability of getting any outcome is same. For tossing a die there are 6 possible outcomes.

Therefore, p = probabiity of gettng any outcome = 1/6.

N = number of times die rolled = 120.

To find expected frequency we use the formula, N*p = 120 * 1/6 = 20.

This expected frequncy is also same for exach face.

Therefore we make a table as follows :

Face value	Frequency	Expected frequency
1	14	20
2	34	20
3	16	20
4	15	20
5	25	20
6	16	20

(a)

From the above table we see that all expected frequncies are greater than 10 which satisfies the condition for runnning chi-square goodness of fit test.

Hence we choose option : Yes, all expeced counts are bigger than 10.

(b)

For null hypothesis we assumes that there is no significant difference between the observed and the expected value. Therefore we choose option : The data fit the distribution for a fair six-sided die.

(c)

Degrees of freedom = Number of E entries -1

Degrees of freedom = 6 - 1 = 5

d)

For this test we use chi square distribution with degrees of freedom 5.

So choose option $\chi ^{2}_{5}$ .

e)

The formula for test statistics is,

$\chi ^{2} = \sum \frac{(Oi-Ei)^{2}}{Ei}$

Oi	Ei	(Oi-Ei)	(Oi-Ei)2	(Oi-Ei)2/Ei
14	20	-6	36	1.8
34	20	14	196	9.8
16	20	-4	16	0.8
15	20	-5	25	1.25
25	20	5	25	1.25
16	20	-4	16	0.8
				$\sum \frac{(Oi-Ei)^{2}}{Ei}$ = 15.7

Therefore,

$\chi ^{2} = \sum \frac{(Oi-Ei)^{2}}{Ei}$ = 15.70

f)

P-value = P( $\chi ^{^{2}}$ > test statistics value)

That is, P-value = P( $\chi ^{^{2}}$ > 15.70)

Using chi square distribution table the probability for test statistics value 15.70 with df 5 is 0.0078.

So we can choose option : if HO is false, then there is a chance equal to p-value that the value of the test statistic will be equal to or greater than the calculated value.

g)

i) Here, level of significance $\alpha$ = 0.05.

ii) P-value = 0.0078.

Since, p-value < $\alpha$ we reject null hypothesis.

iii) Since a > p-value, we reject the null hypothesis.

iv) Since we are rejecting null hypothesis therefore we are in favor of alternative hypothesis.

Hence, there is sufficient evidence to conclude that the data do not fit the distribution for a fair six-sided die.