Question details

STAT 230 Final Exam
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STAT 230 Final Exam

UMUC

Spring 2013

100 Points

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Use the information below to answer Questions 1 through 4.

Given a sample size of 36, with sample mean 670.3 and sample standard deviation 114.9, we perform the following hypothesis test.

Null Hypothesis                    

Alternative Hypothesis         

 

  1. What is the test statistic?       

670.3-700 ÷ 114.9/√36

= -29.7÷19.15

= -1.55

 

  1. At a 10% significance level (90% confidence level), what is the critical value in this test?  Do we reject the null hypothesis?

 

Table V value of .01 is 1.645

-1.645 < z > 1.645    

 

 z= 1.55 which falls within so we accept.

 

 

  1. What are the border values between acceptance and rejection of this hypothesis?

700 +/- 1.645 (100/√ 36) =

700 +/- 1.645 (100/16.6666) =

700 +/- 1.645 (6.00) =

 

700 + 9.87 = 709.87

700- 9.87 = 690.13

666.87, 733.13

 

  1. What is the power of this test if the assumed true mean were 710 instead of 700?

ZL = 709.87 – 710        -0.13    =   -0.0013       Zl=  .0040

            100                     100

ZU = = 690.13 – 710     -19.87             = -0.1987         Zu = .0753

                   100                          100

 

.0753 - .0040 = .0749      β = 1. - .0749 = .9251

 

Questions 5 through 8 involve rolling of dice.

  1. Given a fair, six-sided die, what is the probability of rolling the die twice and getting a “1” each time?

      (1/6) x (1/6)= 1/36

 

 

  1. What is the probability of getting a “1” on the second roll when you get a “1” on the first roll?

P(A|B) = P(A)

P(B|A) = P(B)

 = P(B|A) = P(B) =  1/6

 

 

  1. The House managed to load the die in such a way that the faces “2” and “4” show up twice as frequently as all other faces.  Meanwhile, all the other faces still show up with equal frequency.  What is the probability of getting a “1” when rolling this loaded die?

 

P(1)+P(2)+P(3)+P(4)+P(5)+P(6) = 1

 2x+1x+2x+1x+1x+1x = 1

 8x = 1

 

 x =      1/8 =  0.125

 

 

  1. Write the probability distribution for this loaded die, showing each outcome and its probability.  Also plot a histogram to show the probability distribution.

P(1)

1x .125

.125

P(2)

2x .125

.25

P(3)

1x .125

.125

P(4)

2x .125

.25

P(5)

1x .125

.125

P(6)

1x .125

.125

 

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