MATH E-306 - Theory and Practice of Teaching Statistics

You may (and should) use Fathom for all activities where calculations are required.

Hand in the solutions to the Homework Activities shown below:

Activity 18-7, 18-8, 18-9, 18-10, 18-15, 18-18, 18-20, 18-21Activity 18-7: Charitable Contributions

a.

Fathom found the 90% CI (.7702, .807).b.

Fathom found the 99% CI (.7598, .8174).

c. Yes, the sample proportion does differ from .75 at the .01 significance level because .75 is not in the 99% confidence interval. Thus, you would reject the null hypothesis H0: = .75 vs. the two-sided alternative Ha: .75, and would decide that .75 is not a plausible value for .d. No, the sample proportion does not differ from .80 at the = .10 significance

level because .80 is in the 90% confidence interval. Thus, you would not reject the

null hypothesis H0: = .80 vs. the two-sided alternative Ha: .80. With 90%

confidence, .80 seems to be a plausible value for .Activity 18-8: Female Senators

a.

Fathom results:

The test statistic is z = -6.8b. The p-value is close to 0; it is extremely rare for z to be less than -6.8.c. No, this significance test has no meaning because you know that women constitute less than half of the entire 2007 U.S. Senate because you have taken a census of the population of interest. You know women make up exactly 16% of the 2007 Senate.Activity 18-9: Distinguishing Between Colas

a. The null hypothesis is that the subject is just guessing and would correctly

identify the different brand of soda 1/3 of the time in the long run. In symbols,

the null hypothesis is H0: = .3333.

The alternative hypothesis is that the subject would correctly identify the different

brand of soda more than 1/3 of the time in the long run. In symbols, the

alternative hypothesis is Ha: > .3333.b. You cant tell whether Randys sample data would necessarily lead to rejecting the

null hypothesis because you dont know how many times he ran the experiment. If he only tried twice and succeeded 50% of the time, this would not be convincing evidence that he is doing better than guessing, but if he tried 500 times and succeeded 50% of the time, this would be very convincing evidence.c.

Fathom found the value z=1.905; the probability of getting a value of z at least this large if the null hypothesis is true is .02. So, at the .05 significance level, reject H0 and conclude that Randy is doing better than guessing.d. Assuming Randys success rate is 50%, the CLT says P( > .46) = .7142 with

samples of size 50.

So, you expect about 71% of random samples to yield a sample proportion greater than .46 if Randys success rate is 50% (therefore more than a 50/50 chance for Randy to do so).

e. The probability will increase because with the larger sample size the standard

deviation will decrease from .0707 to .05. (In the formula for z, the sample size, n, is in the denominator of the denominator, so increasing n will increase z.)

f. The probability will increase because the standard deviation will decrease to .06665 and (more importantly) because .46 will be so much further below 2/3 than it is below 1/2. If the population proportion is even further above .46, the probability of obtaining a sample proportion of .46 or more quickly increases.

Activity 18-10: Voter Turnout

a.

Fathom found the 99.9% CI, (.6524, .7123).b. No, his was a self-reported answer about whether people voted. Those surveyed

may have lied to the interviewer or have forgotten whether they voted. It is unlikely that this confidence interval is reliable.c. No, 49% is not within the confidence interval reported in part a.d. Yes, you are 99.9% confident that this confidence interval succeeds in capturing the proportion of all adult Americans who would claim to have voted in 1996. This would be a very different proportion from those who actually did vote.e. You do not need to conduct the significance test because .49 is not contained in the 99.9% confidence interval. You know that you would reject H0: = .49 vs. Ha: .49 at the .1% level of significance.Activity 18-15: Penny Activities

a. A 95% CI for flips: (.5012, .5127)

A 95% CI for spins: (.4435, .4572)

A 95% CI for tilts: (.6829, .6979)

b. These intervals are so narrow because the sample sizes are so large (which makes

the margins-of-error very small).c. You would reject H0: = .5 vs. Ha: .5 at the .05 significance level for all three situations (flips, spins, and tilts) because .5 is not contained in any of the 95% confidence intervals.d. H0: = .5

Ha: .5Flipped pennies:

z = 2.366

p-value = .018

Since p < .05, reject H0The statistical evidence suggests that the probability of obtaining a head with a flipped penny differs significantly from .5 (at the .05 level of significance).

For spins, z = -14.19

p-value < 0.0001

Since p < .05, reject H0The statistical evidence suggests that the probability of obtaining a head with a spun penny differs significantly from .5 (at the .05 level of significance).

For tilts, z = 46.02

p-value < 0.0001

Since p < .05, reject H0The statistical evidence suggests that the probability of obtaining a head with a spun penny differs significantly from .5 (at the .05 level of significance).

e. No; although you are confident that the probability of obtaining a head with a flipped penny is not .50, the probability is quite close to .50. The confidence interval tells you this probability is somewhere between .5012 and .5127, which for all practical purposes is .50.Activity 18-18: Flat Tires

a. H0: = .25

Ha: > .25b. A Type I error would be concluding that more that one-fourth of the students will choose the right-front tire when, in fact, the proportion is not more than one-fourth.

A Type II error would be failing to realize that more than one-fourth of the students will choose the right-front tire when asked the flat tire question.c. Answers will vary. Here is one representative running of the applet:The simulation indicates that you are virtually certain to recognize that the right front tire is being chosen more than one-fourth of the time. The approximate power is 100%.

d. If the sample size were increased from 100 to 200 and all else remained the same, the power of the test would increase. Of course, in this case, it is impossible for the power to increase beyond 100%.

e. If the significance level were increased from .05 to .10 and all else remained the same, the power of the test would increase. Of course, in this particular case, the power cannot increase because it is already 100%.

f. If the right-front tire were actually chosen 40% of the time instead of 50% of the time, and all else remained the same, the power of the test would decrease. This result is verified by the simulation, which shows that the power of the test decreased to about 93.5% in this case. Activity 18-20: Veterans Marital Problems

a. Let represent the population proportion of all Vietnam veterans who are divorced.H0: = .27 (The percentage of all Vietnam veterans who are divorced is 27%.)

Ha: > .27 (The percentage of all Vietnam veterans who are divorced is more than 27%.)b. A Type I error would be deciding that the Vietnam veteran divorce rate is more than 27% when it is actually no more than 27%.c. A Type II error would be failing to realize that the Vietnam veteran divorce rate is more than 27%.Activity 18-21: Hiring Discrimination

A Type I error would be deciding that there is discrimination (concluding that not

enough African Americans are hired by the Hazelwood School District) when,

in fact, there is no discrimination. A Type II error would be failing to realize that the Hazelwood School District was discriminating (not hiring enough African Americans). In this case, which type of error is more serious is a personal opinion

(false accusations vs. allowing discrimination to continue).

MATH E-306 - Topic 18: More Inference ConsiderationsHomework Solutions

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