ISED
160 NOTES ABOUT HOMEWORK, ANNOUNCEMENTS, ETC.
|
Assigned on: |
BE
SURE TO CLICK ON RELOAD/REFRESH ON YOUR
COMPUTER OR THE CURRENT ADDITIONS TO THE PAGE MAY NOT APPEAR! You may also not see current pages if your computer does not have an up-to-date browser… download a new version or use a library/lab computer. Scroll
down as new assignments are added to the old. New assignments are generally
posted by 2:00 pm of the lecture day unless otherwise noted. |
||||||||||||||||||||||||
|
|
Most
current work is listed first, followed by previous entries: |
||||||||||||||||||||||||
|
Th 12/05 |
12/06
UPDATE: Problem from 12/05 below seems to be fixed. The following should now
be the complete notes for the day. Previous message: (My campus
computer is not behaving (maybe too cold?)… the
below problems are all it will let me upload for now after several hours of
failed attempts. I will try uploading the rest of the notes, hmk due Tuesday, and test examples tomorrow from my
computer on another campus … please check back, and sorry for any
inconvenience!) Answers to some of the
last hmk problems from 10.3: 8. Hypotheses: Ho: m = 80 Hi: m < 80 Level of Significance: alpha = 0.02 Data and calculations: t = (76.9 -
80)/(8.5/sqroot22) = – 1.71 Decision: Classical: For 0.02 of alpha
going in left tail, we find critical t value -2.189 in row df = 22 – 1 = 21. Accept Ho. P-value: In row 21, 1.323 < 1.71 < 1.721, so 0.05 <
p < 0.10. Since
p > alpha, accept Ho. Conclusion: No sentence
of meaning since this was not a word problem. 10. Hypotheses: Ho: m = 4.5 Hi: m > 4.5 Level of Significance: alpha = 0.10 Data and calculations: t = (4.9
– 4.5)/(1.3/sqroot13) = 1.11 Decision: Classical: For 0.10 of alpha going
in right tail, we find critical t value of + 1.356 in row df
= 13 – 1 = 12. Accept Ho. P-value: In row 12, 1.083 <
1.11 < 1.356, so 0.10 < p < 0.15. Since p > alpha, accept Ho. Conclusion: No sentence
of meaning since this was not a word problem. 24. Hypotheses: Ho: m = 7.0 Hi: m not equal
to
7.0 Level of Significance: alpha = 0.05 Data and calculations: t = (7.01
– 7.0)/(0.0316/sqroot14) = 1.18 Decision: Classical: For 0.025 of alpha going
in each tail, we find critical t values of +/-2.160 in row df = 14 – 1 = 13. Accept Ho. P-value: In row 13, 1.079 <
1.18 < 1.350, so 2(0.10) < p < 2(0.15), i.e., 0.20 < p < 0.30.
Since p > alpha, accept Ho. Conclusion: The engineer has no evidence of
a bias in the ph meter from the stated 7.0 value. LECTURE/READING: (Read section 11.1)
LAST NEW MATERIAL! We took a look at 11.1 p522 Ex. 2. The matched pairs problems are
still t tests as in 10.3, but allow you to compare two sets of data by
looking at differences between the data. For H0 assume
there is no difference between the data: Ho: m = 0. For H1 there
is a difference of some kind, either m < 0, m > 0, or m is not equal to
0 depending on the phrasing of the word problem and the order in which the
differences are found. Matched pairs
word problem example: An agricultural field
trial compares the yield of two varieties of tomatoes for commercial use. The
researchers divide in half each of 11 small plots of land in different
locations (half gets variety A and half gets variety B) and compare the
yields in pounds per plant at each location. The 11 differences (variety A
minus variety B) give an average of 0.54 and std. deviation of 0.83. Is there
evidence at the 0.05 level of significance that
variety A has a higher yield than variety B? (Assume differences computed by
A yield minus B yield). Answer: Ho: m = 0 If we want to show that plot A tomato yields are
larger, H1 depends on how the
differences are computed. If we subtract the yields in order of plot A
– plot B, then H1: m
> 0 because larger numbers minus smaller numbers will give a positive
number (> 0). But if we subtract the yields in order of plot B – plot A, then H1: m
< 0 because smaller numbers minus larger numbers will give a negative
number (< 0). So in this case, H1: m > 0 We compute sample t=2.16. Using row n-1=10 of the t table
with right tail area of 0.05 the critical value is 1.812 and the estimate for
the p value is 0.025 < p < 0.05 since 1.812 < 2.16 < 2.228.
Either way, we reject the null hypothesis. We have found evidence that
variety A has a higher yield than variety B. HOMEWORK due Tuesday 12/10 (for each matched pairs t-test
below, write hypotheses, calculations, decisions and sentences of conclusion): 1. 11.1 p516 #14c only, given n=6, the mean of the differences
(blue-red) is 0.093, s = 0.17. 2. 11.1 p519 # 20 given differences are computed by “Thrifty minus
Hertz” and n = 10, the mean of the differences is -0.259 and s = 9.20. 3. The design of controls and instruments affects how easily people
can use them. A student project investigated this effect by asking 25 right
-handed students to turn a clockwise screw handle (favorable to
right-handers) with their right hands and then turn a counterclockwise screw
handle (favorable to left-handers) again with their right hands. The times it
took for each handle were measured in seconds, and the 25 differences
(clockwise minus counterclockwise) gave an average of -13.32 seconds with
std. deviation 22.94. Is there evidence that right-handed people find the
clockwise screw handle easier to use? Test at the 0.01 level. Test #5 is a mandatory part of your grade that
will occur as scheduled on Thursday 12/12 and is the last graded activity of
the semester (if you miss it, you will be offered the opportunity to take a
final exam to replace it – this is the only instance in which the final
will be used). It will consist of one page of short answer questions (from
10.1, 10.2, 10.3, 11.1 about use of the z table for p -values, the t table
for critical z values, and the t table for t tests with p-value estimates,
and forming of hypotheses and sentence writing), and one page with 2 word
problems (z test, t test) to perform a complete significance test. Some Review Points: z-tests
use row z on the t-table for critical values and the old z table for p
values: Classical method for z tests: If you have a one-sided z test with alpha = 0.01, n = 35 and sample z = 2.45, on the t table, match the bottom row z (that says z to the left) with the 0.01 column to find critical z value of 2.326. Reject Ho. If it had been a two-sided test, take half of the alpha
(0.01/2 = 0.005) to put in each tail and look at 0.005 column
in row z to find 2.576. Accept Ho. P-value method for z tests: If you again have the one-sided z test with alpha = 0.01,
n = 35 and sample z = 2.45, go to the old z table and put row 2.4 with column
0.05 to find z of the t table and find p = 0.0071. Since the p < alpha,
reject Ho. If it had been a two-sided test, you would have p =
2(0.0071) = 0.0142. Then the p > alpha, so accept Ho. t-tests
use df = n-1 row on the t-table for both the
critical and p value methods: Classical method for t tests: If you have a one-sided t test with alpha = 0.01, n = 35
and sample z = 2.45, go to row n-1 = 34 and find the critical value of 2.441.
Reject Ho. If you have a two-sided test instead, you would go to the
same row, but to the 0.005 column to find the
critical value of 2.728. Accept Ho. P-value method for t tests: If you have a one-sided t test with alpha = 0.01, n = 35
and sample z = 2.45, you look for 2.45 in row 34, and can only find the two
closest values to estimate it: 2.441 < 2.45 < 2.728, so 0.005 < p
< 0.01, p < alpha. Reject Ho. If the test had been two-sided, double the areas to find
0.01 < p < 0.02, so p > alpha. Accept Ho. Some short-answer questions for practice: Example: Find the critical value in a one-sided z
test, n = 45, sample z = -2.59, alpha 0.01. Answer: 2.326 (note that you did not need the n, but
it was given anyway!). Example: a. What is the critical value for a
two-sided t-test with n=33 sample t – 3.02 and alpha of 0.01, , and b. do you reject or accept the null hypothesis? Answer: a.With
row 32 and upper tail of 0.005 since half of alpha goes in each tail, the
critical values are + and – 2.738, so b. we would reject the null
hypothesis. Example: Find the p value for a one-sided z-test, n =
23, sample z= 1.15 and alpha = 0.10. Answer: Since this is a z value problem, you do not
use n on the t table, even though n is given! Go to the old z table and find
the area for z = 1.15, which is p = 0.1251. Example: In the previous example, would you accept or
reject the null hypothesis? Answer: p > alpha, so accept. Example: Estimate for the p value for a one-sided
t-test with n=31 and sample t= 0.65. Answer: In row 30, our t is off the table to the
left, so we know that the p value is larger than the 0.25 that is above the
last table entry, so p > 0.25. Example: What if the previous example had been
two-sided? Answer: We would double the area to account for both
right and left tails, so p > 2(0.25)=0.50. Example: In the previous example, would you accept or
reject the null hypothesis? Answer: the p value is common,
we would accept the null hypothesis no matter what alpha was. Example: Estimate the p value for a two-sided t-test,
n=24, sample t=-2.48, alpha is 0.02. Answer: In row 23, it puts us between columns 0.01
and 0.02 which we must double because it is two-sided, so 0.02 < p <
0.04. Accept the null hypothesis since p > alpha. Example: Write the hypotheses and sentence of
conclusion only for the following z test: The average
score on the SAT Math exam is 505. A test preparatory company claims that the
mean scores of students who take their course is higher than 505. Suppose we
reject the null hypothesis. Answer: Ho : M=505 Hi :
M>505. The company has evidence students who take their course will on
average have a higher score than the 505 of all students who take the SAT
Math exam. Example: (Note the similarities and differences to
the last example)Write the hypotheses and sentence
of conclusion only for the following matched pairs test: A test
preparatory company claims that students planning to take the SAT Math exam
will significantly improve their scores with their test prep system. They
select a random group of students to take the SAT Math exam before taking the course, and
then again, after the course. Differences between the matched data are found
by "After - Before". Suppose we accept the null hypothesis. Answer: Ho : M = 0 (assume the course made no difference) Hi :
M > 0 (the expectation is that the after scores will be higher so
"After - Before" will be a positive number). The company has no
evidence students who take their course will on average have higher scores on
the SAT Math exam. |
||||||||||||||||||||||||
|
T 12/03 |
Note:
My computer malfunction wiped out this assignment on 12/05. The following is
hopefully a close recreation of it! The methods of 10.3 are the
same as those from 10.2 with one exception. Since you must rely completely on
your sample for the std. deviation (instead of having the population std.
deviation from previous studies) you must use a different row than the z row
at the bottom of the table. To use the new part of the t
table, you take one less than the sample size, df=n-1. EXAMPLES
using the t-table: 1. What is the
critical value for a one-sided test with n=20 and alpha =0.05? ANSWER: df=20-1=19 and that row with the column of 0.05
gives a critical value of 1.729 2. What would the
critical value be for the above situation if it were two-sided? ANSWER: In
the same row df=19, you would look at the column
with area 0.025, since half of the alpha of 0.05 goes into each tail, and
this would give you a critical value of 2.093. 3. Find an
estimate for the p value for a one-sided test with n=8 and sample t value of
1.15. ANSWER: df=8-1=7 and in
that row, 1.119 < 1.15 < 1.415 so p value is between 0.10 and 0.15. 4. Find an estimate for the
p value in a one-sided test with n=33 and sample t=0.52. ANSWER: df=33-1=32, so we
look in that row on the new t table to find the next higher and lower numbers
with respect to 0.52. But since 0.52 < 0.682 the p value then is greater than
the area of 0.25 for the t value of 0.682. That is, p > 0.25. 5. Find an
estimate for the p value for a two-sided test with n=25 and sample t value of
1.52. ANSWER: df=25-1=24 and in
that row, 1.318 < 1.52 < 1.711 so the right or left tail area for the p
value is between 0.10 and 0.05, but we have a two-sided test so we double the
areas to get the sum of the left and right tail areas: 0.10 < p < 0.20. 6. Perform a complete t test: To find
out if it seems reasonable that the local town library is lending an average
of 4.2 books per patron, a random sample of 13 people was taken and yielded
an average of 4.75 with std. deviation of 1.65 books. Test at the 0.10 level. Answer: The alternate hypothesis is that m Classical method: Using row n-1=12 of the t table with
tail areas of 0.05, critical value is 1.782 p-value method: estimate for p is 0.20<p<0.30
(double tail areas) since 1.083<1.20<1.356. Accept Ho. We have not found any evidence that the
library is lending a different avg. number of books than 4.2 per person. HOMEWORK DUE Thursday 12/05 10.3 p487 #6, 10,
12, 16, 24 (Skip a. For b use pop mean 7, sample mean 7.01 and s = 0.0316) |
||||||||||||||||||||||||
|
Th 11/21 |
You have all next week off for Thanksgiving Break - Enjoy! READING: 10.3 p481-484 use of new table for t tests. LECTURE – We mainly worked on 10.2,
but I passed out a copy of the following table that we will need for section
10.3. In 10.2, you perform z tests because you have the population std.
deviation. In 10.3, you perform t tests, where you don’t have the population
std. deviation, so you must rely completely on your sample with its sample
std. deviation s. You have the new table to allow for extra error in using
your sample s. The smaller the sample, the more error is allowed for. The “df” in the upper left corner is “degrees of freedom”
which is n – 1, where n is sample size.
The top row and the bottom row
have the numbers you were using in the abbreviated table for looking up
critical values for z tests.
The new t tests will run the
same way as the z tests, just a different row will be used to find values.
For right now, we will just use critical values for t tests. Use the table symmetrically
so that it works for negative values with areas in the left tail. To perform
the p value method, the best we can do is to make estimates. We will talk
about this after your break. EXAMPLES
using the new t table: 1. What is the
critical value for a one-sided z test with alpha =0.05? ANSWER: 1.645 2. What is the
critical value for a one-sided t test with n = 12 and alpha =0.05? ANSWER: For
n = 12, df = 11, so row 11 with column 0.05 gives
1.796 3. What is the
critical value for a two-sided z test with alpha =0.05? ANSWER: With
tail area 0.025, half of the alpha of 0.05 goes into each tail, this would
give you critical values of + or - 1.960. 4. What is the
critical value for a two-sided t test with n = 12 and alpha =0.05? ANSWER: With
each tail area 0.025, row 12 with column 0.025 gives + or – 2.179. Homework due Tuesday 12/03: 1. section 10.2 p477 #22 (skip part a) perform a complete z
test (hypotheses, level, z calculation, decision, sentence) 2. section 10.2 p478 #26 (skip part a, use 6.3 for the sample
mean) perform a complete z test (hypotheses, level, z calculation, decision,
sentence) 3. section 10.2 p478 #28 (skip part a) perform a complete z
test (hypotheses, level, z calculation, decision, sentence) 4. section 10.3 p487 #6 (same as with z tests, but use
appropriate row on new t table) 5. section 10.3 p487 #7 (same as with z tests, but use
appropriate row on new t table) |
||||||||||||||||||||||||
|
T 11/19 |
Grades: Check the 4 test
grades I have for you at http://www.smccd.edu/accounts/callahanp/test160F13.html and
let me know if there are any mistakes. The best 3 of 4
test average is a good starting place for your grade, and those who have been
doing regular hmk and coming to class usually find
that the hmk/quiz grade will enhance the test score
grade by at least half of a grade. This is the best estimate of your grade
for now. NOTES FOR SECTION 10.1: How to state the hypotheses from word problems and
write the sentence of conclusion. HYPOTHESES: H0 is
what is accepted as true for the population mean until evidence to the
contrary is found. H1 is
what the investigator or researcher is trying to show. 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. what was the proposed outcome in alternate hypothesis? If you are
rejecting H0 you have found evidence against H0 and
therefore evidence for H1. If you are
accepting H0, you have not found evidence against H0 and
therefore have not found evidence to back up your claim H1. Write
sentences from the H1 standpoint. SECTION 10.2 Word problems: Do what was in
10.1, but throw the LEVEL of significance, CALCULATION of sample z, and the
DECISION to accept/reject Ho into the middle! 10.1
Example: A Muni bus drives a prescribed
route and the supervisor wants to know whether the average run arrival time
for buses on this route is about every 28 minutes. Suppose that after we calculate
the sample z value the data causes the supervisor to accept H0. Write the hypotheses and the
sentence of conclusion. Answer: H0: M = 28 H1 : M is not
equal to 28 The supervisor has found no
evidence that the average run arrival time for buses on this route is
significantly different from 28 minutes. 10.1
Example: A manufacturer produces a paint which takes 20 minutes to dry. He wants make changes
in the composition to get nicer colors, but not if it increases the drying
time needed. Suppose that after he calculates the sample z value the data
causes him to reject H0. Write
the hypotheses and the sentence of conclusion. Answer: H0 : M = 20 H1 : M > 20 The manufacturer has found
evidence that the composition change significantly increases the drying time,
so he will not make a change. (Notice that he is using the test to pull him
away from a bad decision). 10.2
Example (whole test): According to the Highway
Administration, the mean number of miles driven annually in 1990 was 10,300.
Bob believes that people are driving more today than in 1990 and obtains a
simple random sample of 20 people and finds an average of 12,342 miles.
Assuming a std. deviation of 3500 miles, test Bob’s claim at
the 0.01 level of significance. Hypotheses: Ho: population mean = 10300 Hi: population mean > 10300 Level of Significance: alpha = 0.01 Data and calculations: Z=(12342-10300)/(3500/sqroot20)=2.61 Decision: Classical: alpha of 0.01 gives
a critical z value of 2.326 so reject Ho. P-value: on the z table, 2.61
gives p = 0.0045 so p < alpha. Reject Ho. Conclusion: Bob has found significant evidence
that people are driving more today than in 1990, when they drove an average
of 10,300 miles. 10.2
Example (whole test): Before shopping for a used Corvette,
Grant wants to determine what he should expect to pay. The blue book average
is $37,500. Grant thinks the price is different in his area, so he visits 15
area 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: population mean = 37500 Hi: population
mean not equal to 37500 Level of Significance: alpha = 0.10 Data and calculations: z = (38246.90
- 37500)/(4100/sqroot15) = 0.71 Decision: Classical: For 0.05 of alpha
going in each tail, we find critical z values of +/-1.645. Accept Ho. P-value: For z of 0.71 on the
z table p = 2(0.2389) = 0.4778. Since p > alpha, accept Ho. 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. HOMEWORK
(due Thursday 11/21): 1. 10.1 p461/462 do pair of
problems #16 (state hypotheses) and 24 (write sentence), 2. 10.1 p461/462 do pair of
problems #18 (state hypotheses) and 26 (write sentence), 3. 10.1 p461/462 do pair of
problems #20 (state hypotheses) and 28 (write sentence), 4. Do complete
test (hypotheses, level, calculation, decision, sentence): A researcher
believes that the average height of a woman aged 20 years or older is greater
now than the 1994 mean of 63.7 inches. She obtains a sample of 45 woman and
finds the sample mean to be 63.9 inches. Assume a population std. deviation
of 3.5 inches and test at the 0.05 level. 5. Do complete
test (hypotheses, level, calculation, decision, sentence): The average daily volume of Dell computer stock in 2000 was 31.8 million shares. A trader wants to know if the volume
has changed and takes a random sample of 35 trading days and the mean is
found to be 23.5 million shares. Using a population std. deviation of 14.8
million, test at the 0.01 level of significance. |
||||||||||||||||||||||||
|
Th 11/14 |
Sorry, but I will not be able to post
testscores Friday due to other pressing obligations.
I will now attempt to post by Saturday afternoon! Please check back. Will
attempt to post today’s test scores on this site’s “testscores”
link by Friday afternoon. LECTURE NOTES: Last time, we went
over the hmk problems but added the knowledge of
how to calculate the sample z value and how to perform the significance test
using p-values. A z calculation is on p464 and p-value tests are performed in
ex3 p470 and ex4 p472. The p area is the
probability that you would get a value as far away or farther away from the
mean as the sample value you got. If p < alpha, reject Ho and if p >
alpha, accept Ho You will need to
refer to your old z-table to find the p-values. Abbreviated table
of critical values for reference with the classical approach:
ONE-SIDED EXAMPLE: H0: m =
45 , H1: m < 45 , alpha = 0.05 sample
data: n = 24, sample mean =
40.8, population std dev. = 10.5 z
= (40.8 – 45) divided by (10.5 / sqrroot 24)
= – 1.96 Classical The critical value for 0.05 is –1.645 and –1.96
is farther out, so reject H0.
P-value For z = –1.96 the p
area is 0.0250 using the z table. Since p is < alpha, we reject H0 since the sample is more rare in
probability of occurrence than the alpha. ONE-SIDED EXAMPLE: H0: m =
0.045 and H1:
m > 0.045 and alpha = 0.005 sample z= 2.06 sample
data: n = 17, sample mean =
0.0055, population std dev. = 0.001 z
= (0.0055 – 0.005) divided by (0.001 / sqrroot
17) = 2.06 Classical The critical value for 0.005 is 2.576 and 2.06
is closer to center, so accept H0. P-value The area to the right of 2.06 is 0.0197 so the
p value is 0.0197 > 0.005 (p>alpha) so accept H0
since the sample is not as rare in probability of occurrence as the alpha TWO-SIDED EXAMPLE: H0: m =
35 , H1: m is not 35 , alpha = 0.05 sample
data: n = 40, sample mean =
37.63, population std dev. = 7.4 z
= (37.63 – 35) divided by (7.4 / sqrroot 40)
= 2.25 Classical Each tail gets alpha divided by 2 = 0.025 which has a critical z of 1.96. The sample z
of 2.25 is farther away from the mean than 1.96 so reject H0. P-value For a sample of 2.25, the tail area would be
0.0122 using your old z table. So for a two sided
test, each tail has .0122 so the p value is 0.0122+0.0122=0.0244. Since
0.0244 < 0.05 so we reject H0. MORE EXAMPLES of p value method only: 1. Ho: m=11 and
H1: m not
equal to 11 and alpha =0.01 sample z= 2.67 The area from the
z table for z = 2.67 is 0.0038 and since we are using a two-sided test, we
add the areas for +/- 2.67: p = 2(0.0038) = 0.0076, so p < alpha. Reject Ho. 2. Ho: m=265 and
H1: m < 265 and alpha =0.01 sample z= -2.25 The area from the
z table for z = -2.25 is p = 0.0122, so p > alpha. Accept Ho. 3. Ho: m=35 and
H1: m > 35 and alpha =0.05 sample z= 2.23 The area from the
z table for z = 2.23 is p = 0.0129, so p < alpha. Reject Ho. 4. Ho: m=1.23 and
H1 not equal to1.23 and alpha =0.02 sample z= -2.45 The area from the
z table for z = 2.45 is 0.0071 and since we are using a two-sided test, we
add the areas for +/- 2.45: p = 2(0.0071) = 0.0142, so p < alpha. Reject Ho. 5. Ho m=0.045 and
H1: m > 0.045 and alpha =0.005 sample z= 2.06 The area from the
z table for z = 2.06 is p = 0.0197, so p > alpha. Accept Ho. 6. Ho: m=4500 and
H1: m < 4500 and alpha =0.025 sample z= -1.83 The area from the
z table for z = -1.83 is p = 0.0336, so p > alpha. Accept Ho. HOMEWORK
(due Tuesday 11/19): (Do #1-2 as on
pg.464 and last lecture. You did #3-7 below in the last hmk
using the classical method, now use the p-value method and compare answers -- should have
same conclusions!). 1. Compute z with
n = 35, sample
mean = 36.2, pop. mean = 30,
pop. std dev. =
12.9 2. Compute z with
n = 2500,
sample mean = 24.9,
pop. mean = 25.3, pop. std dev. = 8.4. 3. redo
10.2 p476 #12. given sample z = 1.92, find the
p-value and accept or reject Ho. 4. redo
10.2 p476 #13. given sample z = 3.29, find the
p-value and accept or reject Ho. 5. redo
10.2 p476 #14. given sample z = –1.32, find
the p-value and accept or reject Ho. 6. redo
10.2 p476 #16. given sample z = 1.20, find the
p-value and accept or reject Ho. 7. redo
10.2 p476 #18. given sample z = 2.61, find the
p-value and accept or reject Ho. |
||||||||||||||||||||||||
|
T 11/12 |
No new homework until Thursday
after test. |
||||||||||||||||||||||||
|
Th 11/07 |
LECTURE: Ch 10
Significance tests (10.1 and 10.2): I gave a handout with
the following abbreviated table of critical values, so you do not have to
look them up on the table backwards 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:
One-sided significance test example: If the
null hypothesis is that the mean of a population is 45 and the alternate
hypothesis is that it is less than 45, we have a one-sided alternate hypothesis: you only care if it is less than 45
and you don’t care if it is greater. Ho: m = 35 H1: m < 35 If a level of
significance (alpha) is given as alpha = 0.02 and
you take a sample and standardize it to get z= -1.90, does
it give evidence to reject the null hypothesis and therefore accept the
alternate hypothesis? Classical
approach The critical value comes from the alpha value. Since we only
care about values that stray too far below what is claimed to be the center,
we put all of alpha into the left tail. The critical z value with 0.02 area to its left is -2.054 from the table above. Since
-1.90 is closer to center, we consider it 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. Two-sided significance test example: If the null
hypothesis is that the mean of a population is 35 and the alternate
hypothesis is that it is not 35 (within a certain amount of acceptable
error), we have a two-sided alternate hypothesis: we care if population mean m is significantly higher or lower
than 35. Ho: m = 35 H1: m If a level of
significance (alpha) is given as alpha = 0.05 and
you take a sample and standardize it to get z =2.25, does
it give evidence to reject the null hypothesis and therefore accept the
alternate hypothesis? Classical
approach The critical values come from the alpha value. Since we have a two-sided H1, alpha is divided by 2 to get
0.025 (this is how much goes in each tail of the distribution) and on
the table above, you see a critical z of 1.96. Compare the sample z to
the critical value of z. Since the sample z of 2.25 is farther away from the
mean than the critical value of 1.96, we have evidence to reject the
null hypothesis. MORE EXAMPLES: 1. Ho: m=11 and
H1: m not
equal to 11 and alpha =0.01
sample z= 2.67 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 Ho. 2. Ho: m=265 and
H1: m < 265 and alpha =0.01
sample z= -2.25 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 Ho. 3. Ho: m=35 and
H1: m > 35 and alpha =0.05
sample z= 2.23 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 Ho. 4. Ho: m=1.23 and
H1: m not
equal to 1.23 and alpha =0.02 sample z= -2.45 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 Ho. 5. Ho m=0.045 and
H1: m > 0.045 and alpha =0.005
sample z= 2.06 All of alpha, 0.005, is put into the right tail of the distribution. The
critical value for 0.005 is 2.576 and 2.06 is closer to center than this, so
accept Ho. 6. Ho: m=4500 and
H1: m < 4500 and alpha =0.025
sample z= -1.83 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 Ho. HOMEWORK due Tuesday 11/12: 10.2 p476 (#15-18 in use
p-values, but for right now, treat them with critical values as below). Draw distributions for each
with relevant z values and areas (critical values from table above): 12. given
part a. sample z = 1.92, do parts b, c, d 13. given
part a. sample z = 3.29, do parts b, c, d 14. given
part a. sample z = –1.32, do parts b, c, d 16. given
sample z = 1.20, do part b but using critical values, not p values 18. given
sample z = 2.61, do part b but using critical values, not p values TEST #4 FORMAT for Thursday 11/14.
Given:
the z table, and formulas for the population z value, confidence intervals,
error, and sample size, and the critical values for
90/95/99%. --About 2 word problems with
parts, like 7.3 p354 #17-24, 28-29 and hmks due
10/29, 10/31. --Show how to find the z
critical values for a given confidence level using the z table backwards as
in 9.1 p416 #13-16 (and 7.2 p347 #23-26). --3 or 4 short
answer questions on sampling distributions and the effect of changes to
sample size and confidence on error and confidence intervals, as in hmk due 11/05 and today’s quiz. --At least one
each of word problems (some with follow-up parts) dealing with confidence
intervals, error and sample size (not necessarily in that order), as in 9.1
p416 #21-24, 43-48, and hmk due 11/07. --A few situations
like today’s work to accept or reject hypotheses in significance tests. |
