ISED 160 NOTES ABOUT HOMEWORK, ANNOUNCEMENTS, ETC.
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Assigned on: |
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In
case you lose your z table, here
it is for reference:
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Th 9/13 |
Before starting Test #1, I
asked you to think about using the table backwards to find areas if you are
given a certain amount of area in the middle of the distribution. For instance,
if 80% of the distribution is in the center equally about the mean (40%
directly to the right and left of center), then 10% of the distribution is in
each tail (since 1-0.80=0.20 and half of 0.20 is 0.10). So if you look up
0.10 as an area, you see that the z value that goes with it is 1.28 (positive
and negative values mark each tail on either side of the mean). Make a
picture to see where all of these values are. We call these z values that
mark a predetermined area in the center “critical values”, which I will
denote by z* . We call the area in the center (80% as in the above example)
the confidence level. Work on knowing how to find
the z* or critical values using the z table backwards. As another example,
find the z* for the confidence level of 85%. Answer: Since this area is in
the middle of the distribution about the mean, 1-85% or 15% is outside of
this center range, so 7.5% area is in each tail of the distribution. The
closest area in the table to 7.5% or 0.075 is 0.0749 which gives a z value of
–1.44. So the z* are –1.44 and +1.44. HOMEWORK
(due on Tuesday): Find
the critical values for the following situations (make and label a picture
for each): 1.
Where you have a confidence level of 90% (% of data centered about the mean), what are the z* or
critical values? 2.
Where you have a confidence level of 95% (% of data centered about the mean), what are the z* or
critical values? 3.
Where you have a confidence level of 99% (% of data centered about the mean), what are the z* or
critical values? |
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T 9/18 |
We
have been dealing with normal (bell-shaped) distributions so far. We now
tackle the question of what to do when you do not have a normal population. A
sample is a portion of the population, usually chosen to be as unbiased and representative
of the population as possible. Many times it is too inconvenient or costly to
poll a whole population, so we look for the mean of the smallest sample we
can use that best approximates the population mean without giving too much
anticipated error. For samples, z=(x-` For populations,
z=(x- Now the nice part!
If we look at a population and decide on a sample size n, then we can form a
new distribution (the SAMPLING DISTRIBUTION) made up of all possible samples of that size n
that can be taken from the population. (There are lots of possible samples,
even when the population is relatively small!). Each sample of size n has its
own mean and std. deviation, which are probably not equal to those of other
samples or to those of the population itself. But what can be shown (not in
this class because it involves too much theory and calculation) is that if
you take the average of the means from every possible sample (that is, a mean
of the means!), it will be exactly equal to the mean of the population.
Further, the SAMPLING DISTRIBUTION made up of all the possible samples of
size n will be NORMAL in shape, despite the fact that the population it came
from may not have been normal. So you see, the work you have done so far to
learn how to navigate around the normal distribution has not been in vain!
And if you were worrying about how you were going to find the area under the
curve for non-normal distributions, worry no more. However, there is a
price to pay for such a beautiful result (isn’t there always?). The standard
deviation of the sampling distribution is not the same as the standard
deviation of the population. It varies based on how large a sample size n
that you chose in the first place. The std. deviation of the sampling
distribution is
It
is the same relationship we have been using up to this point, but: 1. the particular
value is not just any x, it is the
mean of a particular sample We will talk (for
sampling distributions), about confidence, error, and sample size. These
topics can be found in any basic statistics book. OUR FORMULAS ( the first
and therefore the second of which I derived in class...the third, which is
the sample size formula, can be derived from the error by solving for n):
Confidence Intervals give
an estimate of the population mean given a particular sample mean, and take
into account how confident you are in using a particular sample average For 90% confidence use z*=1.645,
for 95% use z*=1.96, for 99% use z*=2.576 In
class, we looked at the following problem: A study discloses that 100
randomly selected readers devoted on average 126.5 minutes to the Sunday
edition of the paper. Similar studies have shown a population standard
deviation of 26.4 can be used. Construct a 95% confidence interval for the
true average number of minutes that readers spend on the Sunday edition. The interval, using the
formula above, was 126.5-1.96( 126.5-5.17< Look at the following and
note the changes to the interval: 1.
If we take a smaller sample size, say n=25, then see how the error grows to
10.35: 126.5-1.96( so that 126.5-10.35< 2.
If we take a larger sample size, say n=350, then 126.5-1.96( so that 126.5-2.77< 3.
How about if we go back to the original problem, and insist on a larger
confidence level, say 99%, then 126.5-2.576( so that 126.5-6.80< See how the interval is now
wider, and thus not as helpful an estimate! HOMEWORK due Thursday (will be collected): For each of the following,
use the original in-class example about newspaper reading, 1. Change 95% confidence to
90% confidence and note the effect on the error and the width of the interval
(repeat what we did in class with a new z*). 2. Change 95% confidence to
80% confidence and note the effect on the error and the width of the
interval. Use z*= 1.28 from previous examples. 3. Starting with the
original problem and the 95% confidence level, change the sample size to 1000
and note the effect on the error and the width of the interval. 4. Starting with the
original problem and the 95% confidence level, change the sample size to 40
and note the effect on the error and the width of the interval. 5. In general, from looking
at the examples and your work: a. Does the error get bigger or smaller as you
reduce sample size? b. Does the confidence interval get wider (less
exact estimate) or narrower (closer estimate) around c. Does the error get bigger or
smaller as you reduce the confidence level (%)? d. Does the confidence interval
get wider or narrower around |
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Th 9/20 |
Last homework’s answers: 1. 122.16 to 130.84 2. 123.12 to 129.88 3. error is 1.64 so
124.86 to 128.14 4. error=8.18 so 118.32 to
134.68 5. a. bigger, b.
wider c. smaller d. narrower We talked about the
dynamics of the situations involving confidence and error: THE MORE CONFIDENT YOU WISH
TO BE, THE LESS YOU HAVE TO BE CONFIDENT OF! You might have been thinking
about why you wouldn’t just want to always pick the highest confidence level
possible (like 99.9999%!), but if you look at any problem dealing with
confidence intervals, you can see that when you pick a higher confidence level,
you have more error and hence a worse estimate of the population mean.
Because of this trade-off, you try to pick the best level of confidence to
balance how much error in the estimate will still give you a good idea of .
You will find that there is no such trade-off when changing the sample size.
A larger sample size gives you a tighter interval about the population mean
without decreasing confidence in that interval. But the drawback is that
sometimes you do not have the money or time to allow you to take a larger
sample and you have to settle for the results that you can afford to pay for!
Answers in statistics are not “right” or “wrong” as general mathematical
questions usually are...they are estimates, or educated guesses, that help
you to make decisions. You will have short-answer questions on your next test
which address these dynamics in a general way, and you need to have looked at
problems and thought about how it works before you can answer these questions
properly. I will quiz you on
confidence intervals on Tuesday. Some homework problems to
be turned in are below, and also some additional practice problems if you
wish to try them for yourself (answers are included for the practice
problems). HOMEWORK PROBLEMS TO
BE TURNED IN ON TUESDAY: 1. In problem 1 of the
in-class exercises (where we had a population std. deviation of 12.2 and an
average of 96.4 patient admissions per day), how large a sample of days must
we choose in order to ensure that our estimate of the actual daily number of
hospital admissions is off by no more than five admissions per day? 2. In problem 3 of the
in-class exercises, in thirty-five attempts it took a locksmith an average of
9 minutes to break the code of a certain kind of lock. Assuming that 2.9 is a
reasonable std. deviation from previous attempts, construct a 95% confidence
interval for the true average time that it takes the locksmith to open this
kind of lock. (What has changed from the in-class problem where we had 99%
confidence, and is it a reasonable change for a better outcome?). 3. Suppose that we want to
estimate the average speed of cars traveling on a highway, and we want to be
able to assert with 99% confidence that the error of our estimate, the mean
speed of a random sample of these cars, will be at most 3 miles per hour. How
large a sample will we need if it can be assumed that we can use a std.
deviation of 7.1 miles per hour from previous studies? 4. A random sample of 300
telephone calls made to the office of a large corporation is timed and reveals
that the average call is 6.48 minutes long. Assume a std. deviation of 1.92
minutes can be used. a. What can the office manager say with 99%
confidence about the maximum error if 6.48 minutes is used as an estimate of
the true length of telephone calls made to the office? b. What can the office manager say with 90%
confidence about the maximum error? ********************************* OPTIONAL PRACTICE
PROBLEMS (will not be collected): 1. In problem #1 of the
in-class exercises above, how large a sample of days must we choose in order
to ensure our estimate is off by no more than 2 daily admissions of patients? Answer: z*=1.645, std.dev.=12.2, E=2 so n=101 2. In problem five of the in-class
exercises above, find a 90% confidence interval for the true number of hours
per week that teens watch T.V. if we use a sample of size 111 that yields an
average of 35.8 Answer: z*=1.645, std.dev.=3.2, n=111, x=35.8 so about
35.3 to 36.3. 3. Dr. Paul Oswiecmiski
wants to estimate the mean serum HDL cholesterol of all 20-29 year old males.
How many male subjects would he need in order to estimate the mean serum HDL
of all 20-29 year old males within 1.5 points and using 95% confidence?
Assume that a std. deviation of 12.5 from previous studies is reasonable. Answer: z*=1.96, std. dev.=12.5 and E=1.5 so n=267 4. Given
the word problem take variations of it to see what happens to error and the
interval as numbers are adjusted: A large
hospital finds that in 50 randomly selected days it had on average 96.4
patient admissions per day, assuming a std. deviation of 12.2 from previous
studies. a.
Construct a 90% confidence interval for the actual daily average number of
hospital admissions per day, b.
Construct a 99% confidence interval for the actual daily average number of
hospital admissions per day, c.
Using the 90% interval you constructed, observed what happens to the interval
if you change the sample size to 300, d.
Using the 90% interval you constructed, observed what happens to the interval
if you change the sample size to 20. Note the
effects in general of changing confidence and changing sample size on the
interval width. ANSWERS: a.
96.4-2.84 and 96.4+2.84 b. 96.4-4.44 and 96.4+4.44 c. 96.4-1.16 and 96.4+1.16
d. 96.4-4.49 and 96.4+4.49 5. A Gallup
poll asked 500 randomly selected Americans, "How often do you bathe each
week?". Results of the survey indicated an average of 6.9 times per week.
Using a population std. deviation of 2.8 days, construct a 99% confidence
interval for the number of times Americans bathed each week. Answer: Using the formula for confidence
intervals, n=500, s=2.8, critical z=2.576,we get an error of 0.32 times, so
6.58<m<7.22. In 99 samples out of 100, we would expect that our
estimate of the true number of times that Americans bathe per week is
anywhere from 6.58 times to 7.22 times. 6. In the problem above, how small a sample
can we get away with without being off by more than 1 time bathing in our
estimate? Answer: Using the formula for sample size,
critical z=2.576, s=2.8, E=1, we get n=52.02, so if we ask 53 Americans how
often they bathe, we will be confident 99% of the time that 6.9 times per
week as an estimate of the true number of times that Americans bathe per week
will not be off by more than 1 time. 7. In problem 5 above, what could we say
with 80% confidence about the error in using 6.9 times per week as an
estimate of the true average number of times Americans bathe? Answer: Using the formula for error, n=500,
s=2.8, critical z=1.282, so we get E=0.16. That is, in 80 samples out of 100
we would expect the number of times that Americans bathe per week could be estimated
by 6.9 times and be off by no more than 0.16 times in either direction. (The
less confident you are willing to be in your estimate, the better estimate
you get). Make variations
of the problems above for yourself, such as changing confidence or sample
size, or error, and see what effect the change has on the answers. |
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T 9/25 |
We are now
moving into significance tests. The burden is being shifted from the middle
of the distribution (which was interesting in confidence intervals) to the
tails of the distribution (which are interesting in significance tests). The
middle area, or percentage, was the confidence level. The area in the tails is now referred
to as the significance level, referred to as alpha, or HYPOTHESES: The null
hypothesis, symbolized by The alternate
hypothesis, symbolized by LEVEL OF
SIGNIFICANCE: This is DATA AND
CALCULATIONS: Once a sample
is taken, it will be used to gauge the veracity of the number in the null
hypothesis. To be able to assess this, we must standardize the sample value
so that we can see how many std. deviations it is away from the proposed
center (proposed by the null hypothesis) and therefore be able to look up the
area (or probability of occurrence) associated with it. DECISION: We will then compare
this probability above with the alpha (level of significance) that has been
set, to make our decision about whether or not we believe the null hypothesis
in the light of the evidence that the sample has provided. CONCLUSION: You must state
what you have found from the sample’s evidence, or lack thereof. Write a
grammatically complete sentence with the following elements: Tell 1. if you have
“found evidence” or “not found evidence” against the null hypothesis, 2. about what (what
was the subject of the investigation?), 3. with respect to what number (what was
the number in question in the hypotheses?). *** We will be
doing whole significance tests soon, but first, we must take a more careful
look at each part. We looked at an
example of a full two-sided sigificance test. This is where we care if the
sample is above or below the true population mean. We will look at one-sided
examples next week. An example of a two-sided significance test is as
follows: If the null hypothesis is that the mean of a population is 35 and
the alternate hypothesis is that the mean of the population is not 35 (within
a certain amount of acceptable error) and a level of significance is given as
0.05 (the area outside of the confidence intervals from last time about the
mean) and you take a sample and standardize it to get z=2.25, does it give
enough evidence to reject the null hypothesis and therefore accept the
alternate hypothesis? One way of
determination is to compare the sample z to the critical value of z. The
critical values come from the alpha value. That is, if you look up an area of
alpha (level of significance) divided by 2 =0.025 (this is how much
goes in each tail of the distribution) you see a critical z* of 1.96. So if
you get a sample z that is further away from the mean, you have evidence to
reject the null hypothesis, as in this case. Another way of
determination is to compare the alpha area with the p-value area. The p area
is the probability that you would get a value as far away or farther away
from the center as the sample value you got. For a sample of 2.25, the area
outside of that would be 0.0122. So for a two sided test such as this, just
as the two tails add to alpha in the above problem, you consider that the two
tails would have .0122 in them based on the sample z. Then the p value is
0.0122+0.0122=0.0244. If p<alpha, you reject the null hypothesis and if
p>alpha, then you accept it. Here, 0.0244<0.05 so we again reject
the null hypothesis. The following are two
example problems: Tell using the methods described above if you will accept
or reject the null hypothesis: 1. The alternate hypothesis
is two-sided, 2. The alternate hypothesis
is two-sided, Answers: In the first situation, half of alpha goes into
each tail and the z value that marks the spot such that there is 0.01 area to
the left of it is about –2.325 (or –2.32 or –2.33 if you prefer). Since the
sample z is closer to center than this, we choose to Accept the null hypothesis. If we wished to find the p
value, look up the area for z=2.20 and find 0.0139. The p value is then the
area from the upper and lower tails, that is, 0.0139+0.0139=0.0278. Since
p>alpha, we again Accept the
null hypothesis. In the second situation above, half of
alpha goes into each tail and the z that marks the spot such that there is
0.005 to its left is about –2.575 and +2.575 in the right tail of the
distribution. Since our sample z of 2.73 is farther away from center than
2.575, we Reject the null
hypothesis. The p value for
z=2.73 is 0.0032+0.0032=0.0064, which on its own is a rather small
probability and gives strong evidence against the null hypothesis. But since
we have an alpha of 0.01 and p<alpha, we again have cause to Reject the null hypothesis. I will give you a table of
critical values so that you do not need to look them up using the table
backwards each time you want to perform a test. On Thursday, I will also be
giving problems in class as exercises for you to work on with others and those
who attend will probably feel fairly comfortable with the ideas before they
leave. Those who do not attend will find it difficult to catch up, and will
be confronted with a test on it soon after they show up to class again! So
I strongly suggest that you come on Thursday. There is no formal homework due on Thursday, but to
look at your notes and those I have put above and think about any questions
you have that can assist your learning. HERE IS AN EXAMPLE OF A
WORD PROBLEM TURNED INTO A SIGNIFICANCE TEST (THIS IS WHAT WE WILL DO AFTER
WE BECOME PROFICIENT IN THE METHODS OF ACCEPTING AND REJECTING HYPOTHESES: Grant is in the market to
buy a three-year-old Corvette. Before shopping for the car, he wants to determine
what he should expect to pay. According to the blue book, the average price
of such a car is $37,500. Grant thinks it is different from this price in his
neighborhood, so he visits 15 neighborhood dealers online and finds and
average price of $38,246.90. Assuming a population std. deviation of $4100,
test his claim at the 0.10 level of significance. Hypotheses Ho: Hi: Level of Significance
Data and calculations The sample z comes from the
sample data standardized by the formula:
Next, we determine whether
or not we have sufficient evidence to reject the null hypothesis: Decision: Method 1: For 0.05 of
alpha going in each tail, we find a critical z value of 1.645 and the sample
z is not as far out from the theoretical population mean. Method 2: Look up
sample z of 0.71 on the original z table and find p value of 2(0.2389)=0.4778
> alpha of 0.10. By either method, we find that we do not have evidence to
reject the null hypothesis, so we accept it. Now we form a sentence of
conclusion for what we have found: Conclusion Grant does not have any
evidence that the mean price of a 3 yr. old Corvette is different
from $37,500 in his neighborhood. |
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Th 9/27 |
Keep in mind
that you have your next test on Thursday 10/4. LECTURE NOTES: I gave a
handout which cannot be reproduced here satisfactorily, recapping the symbols
we are using and their meanings. Also included was the following abbreviated table
of critical values, so you do not have to look them up on the table each time
you want to do a problem. The top row represents the area in either the left
or right tail of the distribution, and the bottom row represents the positive
or negative critical value. Refer to it as you look at the example problems
below:
We tried some
two-sided and one-sided examples of significance tests in class, using both
method 1 and method 2. As the one-sided problems were different than the
two-sided ones we focused on last time, here is an example: Ho: m=45 and the alternate hypothesis is one-sided H1:
m>45 , Conclusion: If you have a one-sided alternate
hypothesis ( that is where m is > or < a number instead of
“not equal to”), you don’t take half of alpha as we did in the two-sided
problems, you put it all into the tail of interest. You do not add two tails
to get the p value, you only use the area that you looked up for the one
tail. In the situation above, we only
care about values that stray too far above what is claimed to be the center.
This is a one-sided test unlike the other two situations in the previous
session’s notes. By method 1, you put all of alpha into the right hand tail.
The z value with 0.02 area to its right is about +2.054 from the table above.
Since 1.96 is closer to center, we got a routine sample (one that would
happen 98% of the time) so there is nothing strange about the center being
where it is claimed to be. We Accept the null hypothesis. By method 2, the p value for z=1.96 is 0.0250
using the table.Since p is > alpha, we again choose to Accept the null hypothesis. **************** Things to do
for homework before Tuesday: 1. Read the
solutions to the additional In-class problems below and make sure you understand
them. 2. Read the Example
problems below
with answers and see if you understand them (or if I have inadvertently made
any mistakes, it is your job to find them and talk about them on Tuesday).
These will not be collected, but are for your own benefit to practice. 3. Do the Homework
problems below
(with pictures of the distributions as in class) and they will be collected and talked about on Tuesday. 4. Make note
that you will have Test #2 a week from today as scheduled! The format is below. You
should make sure you know how to do each kind of problem and bring questions
on Tuesday. *************** Additional
In-class problems: 1. H0: Answer: Method
1: Half of alpha, 0.025 goes into each tail since the alternate hypothesis is
two-sided. The critical values for 0.025 are + or – 1.960 and 1.75 is closer
to the center than this, so accept the null hypothesis. Method 2:The
area to the left of 1.75 is 0.0401 so the p value is the sum of the left and
right tails, 0.0401+0.0401=0.0802<0.05 alpha , so accept the null
hypothesis. 2. H0:
Answer: Method
1: All of alpha, 0.05 goes into the left tail of the distribution since the
alternate hypothesis only pertains to values <12.2. The critical value for
0.05 is -1.645 and -1.70 is farther from center than this, so reject the null
hypothesis. Method 2: The
area to the left of -1.70 is the p value from the original z table is
0.0446<0.05 alpha so reject the null hypothesis. 3. H0:
Answer: Method
#1: All of alpha 0.01 goes into the right tail of the distribution since the
alternate hypothesis only pertains to values >15. The critical value for
0.01 is 2.326 and 2.43 is farther from center than this, so reject the null
hypothesis. Method #2: The
area to the right of 2.43 is 0.0075 so the p value from the original z table
is 0.0075<0.01 alpha so reject the null hypothesis. More
example problems Since the “not equal to
sign is not printing out well, it is written out for the two-sided problems: 1. H0:
Answer: Half of
alpha, 0.005 goes into each tail since the alternate hypothesis is two-sided.
The critical values for 0.005 are + or - 2.576 and 2.67 is farther away from
center than this, so reject the null hypothesis. The area to the
left of 2.67 is 0.0038 so the p value is the sum of the left and right tails,
0.0038+0.0038=0.0076<0.01 alpha , so reject the null hypothesis. 2. H0:
Answer: All of
alpha, 0.01 goes into the left tail of the distribution since the alternate
hypothesis only pertains to values <265. The critical value for 0.01 is
-2.326 and -2.25 is closer to center than this, so accept the null
hypothesis. The area to the
left of -2.25 is the p value 0.0122>0.01 alpha so accept the null
hypothesis. 3. H0:
Answer: All of alpha
0.05 goes into the right tail of the distribution since the alternate
hypothesis only pertains to values >35. The critical value for 0.05 is
1.645 and 2.23 is farther from center than this, so reject the null
hypothesis. The area to the
right of 2.23 is 0.0129 so the p value is 0.0129<0.05 alpha so reject the
null hypothesis. 4. H0:
Answer: Half of
alpha, 0.01, is put into each of the left and right tails of the distribution
since the alternate hypothesis pertains to values not equal to 1.23, that is,
both greater than and less than 1.23. The critical values are + or - 2.326
and -2.45 is farther away from center than this, so reject the null
hypothesis. The area to the
left of -2.45 is 0.0071 and to the right of +2.45 is the same, so the p value
is 0.0071+0.0071=0.0142<0.02 alpha so reject the null hypothesis. 5. H0:
Answer: All of
alpha, 0.005, is put into the right tail of the distribution due to the
alternate hypothesis. The critical value for 0.005 is 2.576 and 2.06 is
closer to center than this, so accept the null hypothesis. The area to the
right of 2.06 is 0.0197 so the p value is 0.0197>0.005 alpha so accept the
null hypothesis. 6. H0:
Answer: All of
alpha is put into the left tail of the distribution due to the alternate
hypothesis. The critical value for 0.025 is -1.96 and -1.83 is closer to center
than this, so accept the null hypothesis. The area to the
left of -1.83 is 0.0336 so the p value is 0.0336>0.025 alpha so accept the
null hypothesis. Homework
problems (due Tuesday): Make a picture for each and do both methods 1 and 2 for each
as we did in the in-class work (the pictures cannot be shown here). The “not
equal to” symbol is not printing well, so it is written out!: 1. Ho: 2. Ho: 3. Ho: ****************** Test
#2 format for
Thursday 10/04 (bring any questions you might have on Tuesday! Lots of
practice problems are above over the last two weeks of notes. ): Problem 1: Know how to find
the z* or critical values using the z table backwards. For example, find the
z* for the confidence level of 82%. Since this area is in the middle of the
distribution about the mean, 18% is outside of this range, so 9% is in each
tail of the distribution. The closest area in the table to 9% is 0.0901 which
gives a z value of –1.34. So the z* are –1.34 and +1.34. Problem 2: you
will have some short answer questions on sampling distributions and the
effect of changes to sample size and confidence on error and confidence
intervals. (For example, what is the effect on the error in using the sample
mean to estimate the population mean if you take a smaller sample size? It
gets bigger). Problems 3, 4,
and 5: you will have at least one each of word problems dealing with
confidence intervals, error and sample size, not necessarily in that order
(formulas will be provided and the critical values for 90,95,99% will also be
provided). Problem 6: you
will have about 2 or 3 situations to determine whether you will accept or
reject the null hypothesis using method 1 of comparing z values (sample z and
critical z). Problem 7: you
will have about 2 or 3 situations to determine whether you will accept or
reject the null hypothesis using method 2 of comparing areas (p value and
alpha). |
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T 10/02 |
We went over
the homework, talked about other problems for the test, and then looked at
two full word problem significance tests to prepare for after your test. Some
old and unclaimed homework is in a folder outside my office so please pick it
up if it will help you to study. Homework is to
study for your test. Look over the last couple weeks of notes above for
examples. |
