Please, kind Statistics Experts, who would like to help me with this question? I

Question

Please, kind Statistics Experts, who would like to help me with this question? It has to use the software Stata!

Thank you!

An ordinary six-sided die has been damaged. It no longer turns up
with 1 to 6 dots with equal probability of 1/6. A large number of rolls
of this die con rms that one dot and four dots appear with probability
1/5; two, three and six dots with probability 1/10; and ve dots appear
with probability 3/10. Let S denote the sum of the dots that show up
when the die is rolled twice.
(a) Write down the probability density (mass) function (pdf) of the
random variable S and its cumulative density function (cdf).
(b) Find the population mean and variance of the random variable S.

(c) Using Stata, generate 100 trials of the experiment in which you roll the damaged die twice for every trial and take the sum of the number of dots that turn up (Hint: the Stata command "set obs 100" will generate 100 observations, and the Stata command "gen u runiform will generate a variable called u whose value is randomly drawn from the interval 0 u 1 with a uniform probability density. If you wanted to emulate a coin toss, you could decide that u 0.5 indicates heads and the rest of the interval tails. In this case, you should divide the 0 to 1 interval into 12 appropriately sized segments (d) Plot the outcome of your 100 trials (Hint: use the "histogram command (e) Compare the sample average and standard deviation of your "syn- thetic" data to the population mean and standard deviation of S (Hint: use the "summarize command to obtain these mean and standard deviation (f) Explain how you would go about verifying that the Central Limit Theorem holds for the sample average of the random variable S Clearly state each step in the process. (There is no need to perform this verification or to use Stata (g) Refer to the population mean you calculated in part (b) as Aus, and to the sample average you obtained in part (e) as is, and consider the hypotheses Ho Au us and H Au Aus. Assuming Ho is true, what is the probability of observing a more extrem sample e average than the sample average calculated in part (e)? What is this probability called (Hint: use your results from part (e) to form the appropriate t-statistic.) (h) an you reject the hypothesis Ho at a 5% significance level? At a 10% significance level? What is the lowest significance level at which you can reject Ho? (i) What is the 95% confidence interval for Au given your estimate?

Explanation / Answer

Solution

Let X = number of dots on the face showing up. Then, from the given data, pmf of X is as follows:

x

1

2

3

4

5

6

pmf = P(X = x)

1/5

1/10

1/10

1/5

3/10

1/10

Now, let X1 and X2 be number of dots on the face showing up in Die 1 and Die 2 respectively. Then, pmf of X1 and pmf of X2 are identical to what is given above and S = X1 + X2. S assumes values 2, 3, 4, ……, 12.

P(S = 2) = P(X = 1 and X2 = 2)

= P(X1 = 1).P(X2 = 1) [Because X1 and X2 are independent.]

= (1/5)x(1/5) = 1/25.

P(S = 3) = P(X1 = 1 and X2 = 2 or vice versa)

= {P(X1 = 1).P(X2 = 2)} + {P(X1 = 2).P(X2 = 1)}

= {(1/5)(1/10) } + {(1/10)(1/5)}

= 2/50

P(S = 4) = P(X1 = 1 and X2 = 3 or X1 = 2 and X2 = 2 or X1 = 3 and X2 = 1)

= {(1/5)(1/10)} + {(1/10)(1/10)} + {(1/10)(1/5)} = 5/100

P(S = 5) = P(X1 = 1 and X2 = 4 or X1 = 2 and X2 = 3 or vice versa)

=2[ {(1/5)(1/5)} + {(1/10)(1/10)}] = 10/100

P(S = 6)

= P(X1 = 1 and X2 = 5 or X1 = 2 and X2 = 4 or X1 = 3 and X2 = 3 or X1 = 4 and X2 = 2 or X1 = 5 and X2 = 1)

=2[ {(1/5)(3/10)} + {(1/10)(1/5)}] + {(1/10)(1/10)} = 17/100

P(S = 7) =

P(X1 = 1 and X2 = 6 or X1 = 2 and X2 = 5 or X1 = 3 and X2 = 4 or X1 = 4 and X2 = 3 or X1 = 5 and X2 = 2 or X1 = 6 and X2 = 1)

=2[ {(1/5)(1/10)} + {(1/10)(3/10)} + {(1/10)(1/5)}] = 14/100

P(S = 8)

= P(X1 = 2 and X2 = 6 or X1 = 3 and X2 = 5 or X1 = 4 and X2 = 4 or X1 = 5 and X2 = 3 or X1 = 6 and X2 = 2)

=2[ {(1/10)(1/10)} + {(1/10)(3/10)}] + {(1/5)(1/5)}] = 12/100

P(S = 9) = P(X1 = 3 and X2 = 6 or X1 = 4 and X2 = 5 or vice-versa)

=2[ {(1/10)(1/10)} + {(1/5)(3/10)}] = 14/100

P(S = 10) = P(X1 = 4 and X2 = 6 or X1 = 5 and X2 = 5 or X1 = 6 and X2 = 4)

=2[ {(1/5)(1/10)}] + {(3/10)(3/10)} = 13/100

P(S = 11) = P(X1 = 5 and X2 = 6 or vice-versa)

=2{(3/10)(1/10)} = 6/100

P(S = 12) = P(X1 = 6 and X2 = 6) = (1/10)(1/10) = 1/100

From the above, pmf and cdf of S are:

s

2

3

4

5

6

7

8

9

10

11

12

p(s)

0.04

0.04

0.05

0.10

0.17

0.14

0.12

0.14

0.13

0.06

0.01

cdf = P(S s)

0.04

0.08

0.13

0.23

0.40

0.54

0.66

0.80

0.93

0.99

1.00

NOTE: SUM OF p(s) must be 1

Mean of S = E(S) = Sum{s.p(s)} = 7.2

Variance of S = V(S) = E(S2) – {E(S)}2

E(S2) = Sum{s2.p(s)} = 58.99

So, V(S) = 7.05

x

1

2

3

4

5

6

pmf = P(X = x)

1/5

1/10

1/10

1/5

3/10

1/10

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