**STAT 230 Final Exam**

**UMUC**

**Spring 2013**

**100 Points**

**---------------------------------------------------------------------------------------------------------------------**

**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 **

- What is the test statistic?

670.3-700 ÷ 114.9/√36

= -29.7÷19.15

= -1.55

- 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.

- 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

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

Z_{L }= __709.87 – 710__ __-0.13__ = -0.0013 Z_{l= } .0040

100 100

Z_{U =} = __690.13 – 710__ __-19.87__ = -0.1987 Z_{u }= .0753

100 100

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

**Questions 5 through 8 involve rolling of dice.**

- 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

- 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

- 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

- 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 |

**Due on:**23 Jul, 2016 11:27:35

**Asked on:**23 Jul, 2016 11:27:35

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