Question

In: Statistics and Probability

For this final problem we will create two samples of dice rolls. First You need to...

For this final problem we will create two samples of dice rolls. First You need to roll two dice in the standard way 25 times and record the results. Then roll two dice in a “lucky” way 25 times and record the results. As you complete this problem use the known standard deviation for the sum of two dice : σ1 = 2.415 and σ2 = 2.415

(a) Write down the results of your rolls and find the average for the first sample, ¯x1, and the ”lucky” sample, ¯x2

x1-x2
10	5
7	5
9	4
3	11
11	9
7	8
6	7
10	11
6	5
11	4
8	8
7	5
10	7
10	7
11	8
8	2
5	4
8	11
11	5
12	6
6	3
3	5
10	7
11	8
7	7

(b) Create a 95% confidence interval for the difference in the averages of the two samples, µ1 − µ2.

(c) Create a hypothesis test to determine whether the average of the second sample is greater than the average of the first sample. Be sure to include:

i. The null and alternative hypotheses

ii. The test statistic (and how it was calculated)

iii. The p-value (and how it was calculated)

iv. Your conclusion with α = 0.05. Does this conclusion mean that the “lucky” rolls land on higher numbers than the regular rolls

Expert Solution

mean of sample 1, x̅1= 8.28
mean of sample 2, x̅2= 6.48
..............

b)

Level of Significance ,    α =    0.05
z-critical value =    Z α/2 =    1.9600   [excel function =normsinv(α/2) ]

std error , SE =    √(σ1²/n1+σ2²/n2) =    0.6831
margin of error, E = Z*SE =    1.9600   *   0.683   =   1.3388

difference of means = x̅1 - x̅2 =    8.28   -   6.48   =   1.800

confidence interval is
Interval Lower Limit= (x̅1 - x̅2) - E =    1.800   -   1.339   =   0.4612
Interval Upper Limit= (x̅1 - x̅2) + E =    1.800   +   1.339   =   3.1388
...................

c)

Ho :   µ1 - µ2 =   0
Ha :   µ1-µ2 <   0

Level of Significance ,    α =    0.05

sample #1   ------->
mean of sample 1,    x̅1=   8.28
population std dev of sample 1,   σ1 =    2.415
size of sample 1,    n1=   25

sample #2   --------->
mean of sample 2,    x̅2=   6.48
population std dev of sample 2,   σ2 =    2.415
size of sample 2,    n2=   25

difference in sample means = x̅1 - x̅2 =    8.28   -   6.48   =   1.8

std error , SE =    √(σ1²/n1+σ2²/n2) =    0.6831

Z-statistic = ((x̅1 - x̅2)-µd)/SE =    1.8   /   0.6831   =   2.6352


p-value =        0.0042   [excel function =NORMSDIST(z)]
Desison:   p-value <α , Reject null hypothesis

so it is conclude

that the “lucky” rolls land on higher numbers than the regular rolls

...............

THANKS

revert back for doubt

please upvote

orchestra answered 2 years ago

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