Row 2: 2500 1250 25 50% 1% Row 3: 10,000 5000 50 50% 0.5% Row 4: 1,000,000 500,000 500 50% 0.05%
3. (a) 50,000
3. (b) a zero or a one
3. (c) False
3. (d) True
3. (e) The EV for percentage of the sample is the percentage of the box,
20%; and the SE for percentage is
so 20% plus or minus 1% is a z-score of plus or minus 1/1.33 = 0.75.
By the normal table, the
probability of a result between -0.75 and +0.75 is about 55%.
3. (f) We aren't given information about the number or percentage of forms
in the group with incomes above $75,000, and since incomes are not normally
distributed, a normal approximation for the distribution of the
box is not reliable.
4. (a) 50,000
4. (b) a gross income
4. (c) True
4. (d) True
4. (e) Normal approximation is not very reliable for incomes, but we'll
try it: The EV for the sum of incomes is 900($37,000) = $33,300,000; and
the SE for the sum is 30($20,000) = $600,000. Thus, $33,000,000 corresponds
to a z-score of -0.5, and by the normal table the probability of
a score above -0.5 is (38 + (100-38)/2)% = 69%.
5. EV for sum = 50(150) = 7500; SE for sum = sqrt(50)*35 = 247; so 4 tons,
or 8000 pounds,
corresponds to a z-score of +2.0. The probability of a score above
that is ((100-95.45)/2)% or about 2%. (That is a dangerously small margin
of error, considering the number of trips an elevator makes.)
6. (ii): The size of the population is effectively irrelevant for the accuracy of the survey since a large number of people live in every state. The accuracy is determined by the number of people in the sample. But we will interview 30 times as many people in California, so we should obtain much better accuracy there.
3. (a) 7/500 = 1.4%; (1-0)(sqrt](7/500)(493/500)])/sqrt(500) = 0.5%
3. (b) The traditional rule says 0.4% to 2.4%; but because the 1.4% is
so small, the book says this "95% confidence interval" is not to be trusted.
On the other hand, the lower end of interval, though very small, is still
positive, so ...
4. EV of percent = 379/500 = 75.8%; SE of percent = (1-0)(sqrt(379/500)(121/500)])/sqrt(500) = 1.9%
5. (a) SE = sqrt((.361)(.639))/sqrt(6000) = 0.6%,
so a 95% confidence interval is 34.9% to 37.3%.
5. (b) SE = sqrt((.952)(.048))/sqrt(6000) = 0.3%,
so a 95% confidence interval is 94.6% to 95.8%.
12. (ii) is better: Whatever percent confidence interval is reflected here (probably 95%), only about that fraction of such surveys will have the correct percentage somewhere within its confidence interval; the remaining ones (probably 5%) will have larger errors.
2. (a) True: SE=2.3/sqrt(500) = 0.1. The SE estimates how far
off the sample average is from the population (box) average.
2. (b) True: A 68% confidence interval is the EV give or take the SE.
2. (c) False: The distribution of the box is different from the distribution
of the average of the box. The box could have any type of distribution (for
example, rolling a die has a uniform distribution since all outcomes are
equally likely), but the average of many draws from the box (our estimate
of the box average) has a normal distribution by the central limit theorem.
3. (a) 8.7 miles; SE = 9.0/sqrt(1000) = .3 miles
3. (b) 8.1 to 9.3 miles.
4. It is tempting to do the same kind of arithmetic as in Exercise 3, but the confidence interval for all people based on this survey is likely to be biased. We did not choose people with a simple random survey, we chose households. If,say, the households in the suburbs (with longer commutes) have more people in them, then the sample of people is biased toward longer commutes. In Exercise 3, only the head of the household was included, so the bias did not appear.
5. EV of percent = 721/1000 = 72.1%; SE = sqrt((.721)(.279)))/sqrt(1000) = 1.4%. So a 95% confidence interval is 69.3% to 74.9%.