Because Patricia's section missed this quiz, this quiz will be waived
for her students who were present that day. There were 7 quizzes
total (Quiz 4 was dropped, and the ones afterwards were incorrectly numbered),
and we will drop the lowest. So Sean's students have 6 quizzes for
a total of 60 points.
Patricia's students quizzes will be converted toa percentage
out of 50, and then they will be awarded that percent out of 60.
So if you scored five 10's, you will be given 5 6's. If you scored
5 5's, you will be scored 30 points rather than 25 points. Sorry
for the trouble, but I think this is the fairest way to correct this inbalance.
Quiz 9a, Tuesday, December 8
On any given day, the balance on each of the checking accounts at a
small
credit union is normally distributed with a standard deviation of $750.
A
random sample of 4 checking accounts has these balances:
1637.33, 1425.26, 2415.66, 1630.92
All amounts are in dollars.
1. Find a 90% confidence interval for the mean account balance.
Because the sD is known, we do a z-interval:
Xbar +- z * sigma/sqrt(n)
1777.293 +- z * 750/2
We find z from the Table below. Because we want
a 90% CI, we want to find the value of z that has P(Z > z) = .05.
(So there's 10% "outside" the interval,
90% inside.) Unfortunatley, the Table below is incorrect;
all of the ">"s should have been "<"s. (Or the amounts shown should
have been "1 - the amount").
So what we look for is the point that has about 5% above
it, and the closest we get is 1.65. (The exact value is 1.645, which
you can get from a
computer.)
The CI is then 1777.293 +- 1.65 * (750/2)
(1158.543, 2396.043) is the 90% CI for the mean
balance.
2. Would a 91% confidence interval be wider or less wide than the 90%
interval?
It would be wider. The price we pay for increased confidence
is decreased precision. For example, we could easily give a 100%
confidence interval,
but this interval would be (- infinity, + infinity) which,
while technically correct, has little practical usefulness.
Here are some numbers you need:
If Z is a random variable from a standard normal distribution:
P(Z > .84) = .7995
P(Z > 1.04) = .8508
P(Z > 1.29) = .9015
P(Z > 1.65) = .9505
P(Z > 1.96) = .9750
P(Z > 2.33) = .9904
Quiz 9b, Thursday, December 10
Suppose a study is conducted to determine the income of single women
in
their 30s. It's known that the incomes of such women are normally
distributed with standard deviation of $8000. A random sample
of 4 such
women gives the following incomes (based on tax filings):
25327.22, 22647.43, 24804.98, 26067.69
All amounts are in dollars.
1. Find a 90% confidence interval for the mean income of single women
in
their 30s.
Because the SD of the population is known (SD = 8000),
the z-interval is used:
Xbar +- z * SD/sqrt(n)
24711.83 +- z * (8000/sqrt(4))
To find z, we want the value that, in a standard normal
distribution, has 5% above it. (Not 10%!) See the solution
for Quiz 9a for a discussion and to see what's wrong with the chart I gave
you. The correct value is z = 1.65 (although we could be more accurate
by using a computer and getting 1.645, but this
value is not available to you while taking an exam.)
(18111.8, 31311.83) is the 90% CI for the mean income of single women in their 30s.
2. Would a 89% confidence interval be wider or less wide than the 90% CI?
Smaller. Because we're willing to be less confident,
we can be more precise. To take an extreme, we could give a 0% confidence
interval as an empty interval (,). This is technically true, but
of no use to anyone.
Here are some numbers you need:
If Z is a random variable from a standard normal distribution:
P(Z > .84) = .7995
P(Z > 1.04) = .8508
P(Z > 1.29) = .9015
P(Z > 1.65) = .9505
P(Z > 1.96) = .9750
P(Z > 2.33) = .9904