Topic 17 Answers

Activity 17-1

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Answers for Activity 17-1

Activity 17-2

a) Graphic below

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)

Activity 17-3

b) no
c) 0.3092
d) 0.1546

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

Activity 17-4

a) Be sure to use the website program: spincoin program
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.

Activity 17-5

a) What do you think? I think it would have to be more narrow
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.

Activity 17-6

a) The larger the sample size, the more narrow the C.I. since the standard
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

Activity 17-8

a) 106/355=0.29859, statistic
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.

Activity 17-10

a) Population - viewers of Nick-at-Nite in June 1994
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

Activity 17-11

a) Z*(0.99) = 2.576. Using the same method as the previous problem
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

Activity 17-13

a) 94 couples had different ages when married
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.

Activity 17-14

a) (0.46656,0.50986)
b) Yes
c) Since our interval contains 0.5 it suggests that half of all American
adults consider themselves moderates

Activity 17-15

a) Check our data, mine unfortunately is at school :-(
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

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