Tail | Tail | Head | Tail | Head | Tail | Head | Head |

Tail | Head | Tail | Tail | Head | Head | Head | Tail |

Head | Head | Tail | Tail | Tail | Head | Head | Head |

Head | Tail | Head | Head | Head | Tail | Head | Tail |

Tail | Head | Tail | Tail | Head | Head | Tail | Tail |

Answers for Activity 17-1

b) Area = .95 + 0.025 = .975 The .025 comes from the left tail. Recall 0.05 of

the total area remains, thus 0.025 at either end of the curve.

c) 1.96 using the invnorm(0.975) on TI-83

d) 1.44 using the invnorm(0.925)

b) no

c) 0.3092

d) 0.1546

e) You are using random sampling techniques and your sample size is at 30 items

b) Varies with your results to part a c) Not likely, but you could get lucky

d) approximately 95% of the intervals generated will contain the actualy parameter value.

b) For TI-83 people, you will find out the the interval grows wider,

as the confidence interval grows. To varify this run the program with various values

c) You can check your guess against what actually happened.

deviation will decrease in size (square root term)

b) You can still do this without the calculator program, just do the problem by hand!

Please check you answers at: Answer for part B

c) Your call

d) twice as big

e) cuts the half-width in half

b) parmeter

c) 90% -> (0.2586,0.3386)

95% -> (0.2510,0.3462)

99% -> (0.2360,0.3612)

d) For an C.I. of 90% we would expect 90 out of 100 intervals to contain the actual

proportion of games decided by one run.

e) Not a true random sample, there could be many extraneous factors involved.

f) Possibly, but still too many factors are not considered.

Parmeter of interest - who had more magical powers Jeannie or Samantha

b) Not random, only those who cared enough to call

d) Huge sample size reduces the standard deviation to a very small value

we find the C.I. will be (0.29907,0.31093)

b) In 100 samples, we would expect 99% of our intervals to contain the

proportion of our sample

c) We are assuming that this is a true random sample

b) 67

c) (0.636,0.78953)

d) (0.6213,0.80424)

e) (0.59244,0.83298)

f) no

g) Since none of our intervals contained 0.50, we can reasonably assume

that more than half the marraiges, have a bride younger than the groom.

b) Yes

c) Since our interval contains 0.5 it suggests that half of all American

adults consider themselves moderates

b) Quadruple the sample size, cuts the width in half

c) Half the answer again, thus the width is one-fourth the original

Go to the Moore textbook answers

Submit your question here. Do not be shy, I will check my mail at least once a day and reply as quickly as possible!!