CHAPTER 10—HYPOTHESIS TESTING
Chapter 10--Hypothesis Testing.Doc
STATISTICS 301—APPLIED STATISTICS, Statistics for Engineers and Scientists, Walpole, Myers, Myers, and Ye, Prentice Hall
Goal: In this section we discuss the “Hypothesis Test” statistical inference. We will investigate how to “test” an hypothesis involving a population mean, proportion, difference of two means, and difference of two proportions.
We note that any hypothesis about a population parameter can be tested. The only requirement is that the distribution of the estimator of the parameter must be known.
We first review what takes place in a courtroom setting in the United States.
US COURTROOM SETTING
Consider what happens in the typical courtroom trial in the US. Suppose I am charged with some crime. What happens?
What is the basic tenet in a trial in the US?
Innocent ‘til proven Guilty. What does this mean or imply?
If I am charged with a crime, who goes first, the prosecution or defense, and what must they show or prove?
Prosecution and must prove “within reasonable doubt” that I committed the crime.
What does “reasonable doubt” mean?
What does the prosecution do to present his/her case? What does the defense do?
Prosecution presents evidence that I committed the crime, while the defense presents evidence to refute the prosecution’s evidence and back and forth.
When both the prosecution and defense are finished with their arguments, the case goes to the jury. What verdicts can the jury return?
Guilty as charged
Not Guilty or Acquittal—Do they mean the same thing?
Hypothesis testing is the same idea in statistics.
1. We will assume a statement about a population parameter is true.
2. We will collect sample data.
3. We will use the distribution of sample statistic, assuming our initial statement about the population parameter is true.
4. We will see where the sample results “fall” in the distribution found in 3) and then determine if our sample is “reasonable” or “unreasonable” given our initial assumption about the parameter.
5. If the null hypothesis is true, our sample results should be something we expect to occur.
The setup for every hypothesis test is identical. The differences lie in the parameter of interest and the distribution of the statistic used. Hence our goal of hypothesis testing is:
Goal: The goal of hypothesis testing is to determine whether a statement about a population parameter (or set of parameters) is reasonable or not.
Our decision will be based on sample evidence in the form of a statistic.
The sampling distribution of the statistic will be used to judge whether or not our statement is reasonable.
Recall that in the previous sections we obtained the sampling distributions of the sample average, sample variance, sample proportion, the difference of two sample averages and two sample proportions.
|Known Estimators & Sampling Distributions |
|( [pic] = Z if popln is Normal and ( known |
| |
|( [pic] |
|( [pic] |
| |
|[pic] |
|( [pic] ≈ Z if np > 5 & n(1-p) > 5 or n[pic] > 5 & n(1-[pic]) > 5 |
|( [pic]≈ Z if n1[pic] > 5, n1(1-[pic]) > 5, n2[pic] > 5, and n2(1-[pic]) > 5 |
TERMINOLOGY
The statement about the population parameter is called the Null Hypothesis and is denoted, Ho.
The “opposite” of the statement concerning the parameter in Ho is denoted as the Alternative Hypothesis and is denoted, HA.
When setting up hypotheses, the usual practice is to make the Alternative Hypothesis the statement about the parameter that you would like to show, then make the Null Hypothesis the opposite of this statement. Since we usually trying to show some effect or some difference, this amounts to making the Null Hypothesis the statement about the parameter that amounts to a “no effect” or “no difference” statement about the parameter or parameters.
The Null Hypothesis is assumed to be true until we can prove it false based on sample evidence.
EXAMPLE #1
For example, suppose the Provost comes to me with the following question. He thinks (and would like to show) that the typical student on the Middletown campus is older than the typical student on the Oxford campus. For some reason he knows that the mean age of students on the Oxford campus is 20.2 years.
If we let (MDL be the mean age of all Middletown students, we need to test an hypothesis concerning (MDL. Note that the Dean wants to know if the mean age of Middletown students is greater than 20.2. This is the hypothesis we wish to prove or show so this becomes our Alternative Hypothesis, so HA: (MDL > 20.2. The Null Hypothesis is the opposite of this statement so it is, Ho: (MDL ( 20.2. Said another way, Ho is also Ho: (MDL = 20.2 or less. Hence our pair of hypotheses would be
Ho: (MDL = 20.2 or less
HA: (MDL > 20.2
EXAMPLE #2
Consider the following question from a different text: The national unemployment rate last month was 4.3%. You think the rate may be different in your city, so you plan a sample survey that will ask the same question as the Current Population Survey.
1. What is the population and parameter of interest?
2. To see if the local rate differs from the national rate, what would your hypotheses be?
SETTING UP HYPOTHESES
So in general, when you write down your hypotheses, they should look like:
Ho: the opposite of the statement in HA.
HA: a statement about the parameter that we would like to prove or show.
Notes and Comments
1. When you set up your hypothesis, the statement about your parameter in Ho MUST contain the “=.” Notice that in each of the hypotheses found thus far this has been the case.
DECISIONS IN HYPOTHESIS TESTING
To decide between the Null and Alternative hypothesis we will use sample evidence. We will:
1. take a random sample, or in the case of more than one population of interest, two or more random samples (usually independent!),
2. obtain the statistic that is an estimate of the parameter of interest ([pic] for (, s2 for (2, [pic] for p, etc,),
3. determine the sampling distribution of the statistic,
4. determine where our observed statistic falls in this sampling distribution by obtaining the “p-value” of the test (where small p-values imply the result is very unusual and large p-values imply the result is not unusual),
the “p-value” is the probability of obtaining a result as, or more extreme, than our sample result,
5. reach a decision about our hypothesis based on the p-value as follows:
If p-value large ( > (), then Fail To Reject Ho.
If p-value small ( ( (), then Reject Ho,
where ( is known as the Level of Significance and is usually set to a value of 0.01, 0.05, or 0.10. The Level of Significance is the probability of making one of two kinds of mistakes in a hypothesis test, so we set it fairly small to insure, somewhat, that we do not make this kind of mistake.
6. and, lastly, interpret our results.
While this list is daunting, you will not need to be concerned about calculating p-values, but you will need to know how to interpret them. How you interpret them is in #5. The rule is “If the p-value is small, then Reject the Null Hypothesis. Otherwise, Fail To Reject the Null Hypothesis.”
Notes and Comments
2. Note that we say “Fail To Reject Ho” and NOT “Accept Ho.” The difference is important in that “Accept” gives the impression that Ho is true. This is not the case and an analogy occurs in trials by jury. In a jury trial, the defendant is assumed innocent until proven guilty, beyond reasonable doubt. This is just like hypothesis testing. Ho is that the person is innocent and HA is the person is guilty. The arguments given by the prosecution and the defense serve as the “sample evidence.” The two possible verdicts are:
1. Guilty ( Reject “Ho: Innocence” in favor of “HA: Guilt”
2. Acquittal ( Fail To Reject Ho.
Note that acquittal does not imply innocence, but rather the evidence was not sufficient to convict or find guilty.
3. When performing a hypothesis test, it is the statistical test that is significant. Always consider the practical relevance of your results. A statistically significant result may not be of any practical significance.
Similarity to the Courtroom Setting
| |US Courtroom Setting |Hypothesis Testing |
|What are we interested in? |A person is of interest. |The parameter of interest. |
|What’s assumed true from the outset? |Person innocent until proven guilty. |Ho is assumed true until proven false |
|Evidence to use in decision. |Evidence presented by prosecution and defense. |Sample evidence and a sample statistic. |
|Decisions are: | | |
|Guilty |Evidence proved “beyond reasonable doubt” the |Sample evidence “proves” beyond reasonable doubt |
| |person committed crime. |that Ho is not true and hence HA is more likely to|
| | |be true. |
| | | |
| | |Our conclusion will be that we “Reject Ho in favor|
| | |of HA.” |
|Acquittal |Evidence was not provided that proved the |Sample statistic result does not give any |
| |person committed crime. |indication that HA is possible. |
| | | |
| | |Our conclusion will be that we “Fail To Reject |
| | |Ho.” |
Hypothesis Testing Steps
Make sure the following statements hold in your hypothesis test.
i. Always put the statement to be shown or proven in HA.
ii. Make sure Ho and HA are complementary statements involving the parameter.
iii. Make sure the = sign is in the null hypothesis.
iv. Assume Ho true until proven otherwise.
The basic steps in ANY!!!! hypothesis testing. That is, every test of an hypothesis is set up in the exact same way!
0. Define the parameter(s) of interest.
1. Ho:
2. HA:
3. Set , the level of significance; typically 0.10, 0.05, 0.01.
4. Obtain a Point Estimator(s) of the parameter(s) and determine the distribution of the statistic(s).
5. Calculate the point estimate and
p-value = Pr { Point Estimator more extreme than observed point estimate }.
6. Draw your conclusion:
[pic] If p-value large ( > (), then Fail To Reject Ho.
[pic] If p-value small ( ( (), then Reject Ho.
7. Interpret results.
EXAMPLE #3
“A chemist measures the haptoglobin concentration (in gm/l) in the blood serum taken from a RS of 8 healthy adults; the values are: 1.82, 3.32, 1.07, 1.27, 0.49, 3.79, 0.15, & 1.98. Is there statistical evidence that the mean haptoglobin concentration in adults is greater than 1 gm/l?
What do we know?
0.
1. Ho:
2. HA:
3. Set =
4.
5.
6.
7.
Below is an article on waiting times in drive-throughs at fast food restaurants from the Cincinnati Enquirer a few years back.
|[pic] |
|[pic] |
EXAMPLE #4
In the previous article about fast foods, let’s determine if the mean time in minutes at a Wendy’s drive through is less than the industry average of 3.38 minutes. A random sample of 600 drive-through times at Wendy’s yields an average of 2.50 minutes with a standard deviation of 1.724 minutes.
What do we know?
0.
1. Ho:
2. HA:
3. Set =
4.
5.
6.
7.
EXAMPLE #4A Wendy Example Revisited
Why do make the statement we want to prove or show HA and not Ho? Let’s redo the Wendy’s example this other way and see what we get. We have filled in the first six steps.
What do we know? Since n large (really large!), [pic] is approx Normal with mean 3.38 and variance 0.00495, while our estimate is still [pic] = 2.50.
0. ( = mean drive thru wait time at Wendy’s restaurants
1. Ho: ( = 3.38 ( or less)
2. HA: ( > 3.38
3. Set = 0.05
4. [pic] is approx Normal with mean 3.38 and variance 0.00495.
5. Our point estimate (statistic) is [pic] = 2.50.
The p-value = Pr{ [pic] > 2.50 } = Pr{ Z > -12.08 } = 1.0000.
6. Since p-value = 1.0000 > α = 0.05, we FAIL TO REJECT Ho.
7. With 95% confidence we can NOT conclude that the mean drive thru time at Wendy’s is greater than 3.38.
Note that this DOES NOT IMPLY that ( is ≤ 3.38!!
So here’s why we make HA: the statement we want to show.
Rejecting a Null in favor of the Alternative is a stronger statement that Failing to Reject a Null!
It’s just like a Guilty verdict compared to an Acquittal!
EXAMPLE #5
Consider the claim that two liter bottles of pop do not contain two liters. Suppose a random sample of seven bottles is taken and the contents of the bottles are determined to be: 1.8, 2.2, 2.4, 1.8, 2.0. 2.2, and 1.6 liters.
What do we know?
0.
1. Ho:
2. HA:
3. Set =
4.
5.
6.
7.
EXAMPLE #6 Hair Color/Pain Threshold
Light Blonde: 62 60 71 55 48 ( nLB = 5, [pic]=59.2, sLB = 8.5264, se([pic]) = 3.8131
Dark Brunette: 32 39 51 30 35 ( nDB = 5, [pic]=37.4, sDB = 8.3247, se([pic]) = 3.7229
What do we know?
0.
1. Ho:
2. HA:
3. Set =
4.
5.
6.
7.
EXAMPLE #7
What do we know?
0.
1. Ho:
2. HA:
3. Set =
4.
5.
6.
7.
EXAMPLE #8
Consider the following article from the New York Times.
[pic]
What do we know?
0.
1. Ho:
2. HA:
3. Set =
4.
5.
6.
7.
USING SAS TO TEST HYPOTHESES
SAS Test for a Single Mean
OPTIONS LS=110 PS=60 PAGENO=1 NODATE;
TITLE 'Testing.SAS';
TITLE2 'EXAMPLE OF ONE AND TWO SAMPLE TEST OF MEAN IN SAS';
TITLE3 'POP DATA FROM CLASS';
DATA POPDATA;
INPUT AMOUNT @@; AMOUNT2=AMOUNT-2;
DATALINES;
1.8 2.2 2.4 1.8 2.0 2.2 1.6
;
PROC PRINT DATA=POPDATA;
PROC TTEST DATA=POPDATA ALPHA=0.05;
VAR AMOUNT AMOUNT2;
RUN;
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Testing.SAS 1
EXAMPLE OF ONE SAMPLE TEST OF MEAN
Obs AMOUNT AMOUNT2
1 1.8 -0.2
2 2.2 0.2
3 2.4 0.4
4 1.8 -0.2
5 2.0 0.0
6 2.2 0.2
7 1.6 -0.4
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Testing.SAS 2
EXAMPLE OF ONE SAMPLE TEST OF MEAN
The TTEST Procedure
Statistics
Lower CL Upper CL Lower CL Upper CL
Variable N Mean Mean Mean Std Dev Std Dev Std Dev Std Err Minimum Maximum
AMOUNT 7 1.7384 2 2.2616 0.1823 0.2828 0.6228 0.1069 1.6 2.4
AMOUNT2 7 -0.262 63E-18 0.2616 0.1823 0.2828 0.6228 0.1069 -0.4 0.4
T-Tests
Variable DF t Value Pr > |t|
AMOUNT 6 18.71 |t|
WGTGAIN Pooled Equal 43 2.49 0.0166
WGTGAIN Satterthwaite Unequal 32.9 2.46 0.0192
Equality of Variances
Variable Method Num DF Den DF F Value Pr > F
WGTGAIN Folded F 21 22 3.11 0.0107
SAS Test of the Difference of Two Means
TITLE3 'BATTERY DATA';
PROC TTEST DATA='C:\MyDocs\Class\STA 301\Data Sets\BATTERY' ALPHA=0.10;
CLASS BATTERYTYPE;
VAR TIME;
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Testing.SAS 3
EXAMPLE OF ONE AND TWO SAMPLE TEST OF MEAN IN SAS
VITAMIN C PAIRED DATA
The TTEST Procedure
Statistics
Lower CL Upper CL Lower CL Upper CL
Variable BatteryType N Mean Mean Mean Std Dev Std Dev Std Dev Std Err
Time LithiumIon 53 13.554 13.826 14.098 1.0199 1.1819 1.4119 0.1623
Time NickelMetalHydride 45 13.83 14.348 14.867 1.765 2.0693 2.5149 0.3085
Time Diff (1-2) -1.078 -0.522 0.0328 1.4757 1.649 1.8731 0.3343
T-Tests
Variable Method Variances DF t Value Pr > |t|
Time Pooled Equal 96 -1.56 0.1214
Time Satterthwaite Unequal 67.4 -1.50 0.1387
Equality of Variances
Variable Method Num DF Den DF F Value Pr > F
Time Folded F 44 52 3.07 0.0001
SAS Test of the Difference of Two Means—Paired Data Case
TITLE3 'VITAMIN C PAIRED DATA';
DATA PAIRED;
INPUT VITAMINC PLACEBO;
DIFFERENCE = VITAMINC-PLACEBO;
DATALINES;
145 417
185 279
387 678
593 636
248 170
245 699
349 372
902 582
159 363
122 258
264 288
1052 526
218 180
117 172
185 278
;
PROC TTEST DATA=PAIRED;
VAR DIFFERENCE;
RUN;
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Testing.SAS 4
EXAMPLE OF ONE AND TWO SAMPLE TEST OF MEAN IN SAS
VITAMIN C PAIRED DATA
The TTEST Procedure
Statistics
Lower CL Upper CL Lower CL Upper CL
Variable N Mean Mean Mean Std Dev Std Dev Std Dev Std Err Minimum Maximum
DIFFERENCE 15 -180.8 -48.47 83.89 174.98 239 376.93 61.711 -454 526
T-Tests
Variable DF t Value Pr > |t|
DIFFERENCE 14 -0.79 0.4453
KINDS OF ERRORS IN HYPOTHESIS TESTING
There are two kinds of mistakes that we can make in hypothesis tests. Recall that either Ho and HA must be true since they are complementary statements.
| | |Reality or Actual Truth |
| | |Ho True |HA True |
|Our Decision |Fail To Reject Ho |OK |Type II or β Error |
| |Reject Ho in favor of HA |Type I or α Error |OK |
α = Pr{ Type I Error } = Pr{ Reject Ho when Ho true }
β = Pr{ Type II Error } = Pr{ FTR Ho when HA true }
Notes:
1. α usually specified as 0.10, 0.05, 0.01, or smaller.
2. Power of a Test = 1 - β = Pr{ Reject Ho when HA true }
3. As you make ( smaller, ( gets larger.
Examples
1. The EPA is concerned that the mean DO content of the river is less than 5 due to pollution by a sewage treatment plant upstream. Hence:
Ho: ( = 5 or (>) ( Plant is not affecting the river DO content
HA: ( < 5 ( Plant is adversely affecting the river DO content
Type I Error is Reject Ho when Ho is true (
Type II Error is FTR Ho when HA is true (
2. Consider a drug company with a new cancer drug.
Ho: New drug is not better than the current drug
HA: New drug is better than the current drug
Type I Error is Reject Ho when Ho is true (
Type II Error is FTR Ho when HA is true (
Comparison of Confidence Intervals and Hypothesis Testing
Many claim that CI’s are more informative than hypothesis testing since they provide information on the value of the parameter AND they can be used to test hypotheses in the following fashion:
If we have a Confidence Interval for a parameter with a certain degree of confidence, we can test a certain hypotheses about this parameter. Namely we can test hypotheses of the following form:
Ho: parameter = value versus Ha: parameter ( value.
Our decision about these hypotheses using the CI would be as follows:
Reject Ho if value NOT in the CI
Fail To Reject Ho if value is ANYWHERE in the CI.
Example
Suppose we wanted to test Mars claim that all “vending sized” Milky Way candy bars weigh at least 58.1 grams. Let p = the proportion of Milky Way candy bars that have a weight under 58.1 grams. We would test Ho: p = 0 versus HA: p ( 0. A 95% confidence interval for p was found to be (4%, 26%). We would Reject Ho in this case since our Null Hypothesis value, 0 in here, is not in the interval.
SPECIAL TESTS IN STATISTICS—16.1: SIGN TEST OF MEDIAN
Goal: Our goal in this section is present the methodology to test the median of a population.
Below is the eight step hypothesis test of a population median. The rationale of the test is based on the definition of the median, which has 50% of the observations on either side. Hence if the population median is actually some value, say [pic], then the number of observations in the sample less than [pic] is a Binomial RV with parameters, n and p = ½.
0. [pic]= Median of a population
| |Lower Tailed | |Two-Tailed | |Upper Tailed |
|1. Ho: |[pic] ( [pic] | |[pic] = [pic] | |[pic] ( [pic] |
| | | | | | |
|2. HA: |[pic] < [pic] | |[pic] ( [pic] | |[pic] > [pic] |
3. Set α
4. Test Statistic = X = Number in the RS < [pic]
5. Let x be the observed number in the RS < [pic] where values = [pic] are ignored and n adjusted accordingly
Calculate the p-value where X ~ Binomial ( n, p = ½ )
|Pr{ X ( x } | |2*Pr{ X ( x } if x < n/2 | |Pr{ X ( x } |
| | | | | |
| | |2*Pr{ X ( x } if x > n/2 | | |
6. Conclusion
7. Interpretation
EXAMPLE #1—Example 16.1 pg 674 (WMMY 8th)
A sample of n = 11 electric trimmers yields the following times (in hours) that a rechargeable trimmer operates before it requires recharging: { 1.5, 2.2, 0.9, 1.3, 2.0, 1.6, 1.8, 1.5, 2.0, 1.2, 1.7}. The company claims a median of 1.8 hours before the trimmer needs recharging.
0. [pic]= Median time (hrs) an electric recharge time
1. Ho: [pic] = 1.8
2. HA: [pic] ( 1.8
3. Set =
4. Test Statistic = X = Number in the RS < 1.8
5. We note that x = 7 are less than 1.8, ignoring the one observation = 1.8 so
X ~ Binomial ( n = 10, p = ½ ).
Hence the p-value = 2*Pr{ X ( 7 } (since 7 > 10(2 = 5) and yields
p-value = 2*( 1 – Pr { X 1, no more than 20% < 5.
p-value = Pr { [pic] > X2 }
6. Conclusion.
7. Interpretation.
NOTES AND COMMENTS
1. Note the assumptions for the “legal” use of the test:
all Expected Counts > 1 AND no more than 20% < 5
2. Note also that the test is “one-tailed” (we only reject if we get something TOO BIG) even though the Alternative Hypothesis is two-tailed!
Example #1: Favorite Numerical Digit
Is the distribution of “Favorite Numerical Digit” an equally likely distribution?
Recall what it means to be equally likely!
Using the data from all of the classes that I have taken the survey, here is the observed frequency of each digit.
Tally for Discrete Variables: Favorite Digit
Favorite
Digit Count
0 4
1 13
2 26
3 55
4 42
5 39
6 26
7 76
8 24
9 23
N= 328
[pic]
What do you think?
Note that your answer is subjective. A test removes the subjectivity and makes our conclusion objective.
Here’s our results in a table format.
|Digit |0 |1 |2 |
|Female |14 |35 |49 |
|Male |35 |39 |74 |
|Total |49 |74 |123 |
CHI-SQUARE TEST OF INDEPENDENCE TEMPLATE
1. Ho: Variables A & B are Independent
2. HA: Variables A & B are Dependent
3. Set α.
4. Calculate test statistic:
X2 = [pic].
5. Calculate the p-value. Assumes n large, i.e., all Cell Exptd Counts > 1 & no more than 20% < 5!
p-value = Pr { [pic] > X2 } where df = (a – 1)(b – 1)
6. Reject Ho if p-value small (< ().
7. Interpret results.
Note that the null hypothesis can also be expressed in terms of the proportions as follows. Let pij be the proportion of items in the population with value “i” of Variable A and value “j” of Variable B so that our table becomes:
| | |B |
| | |1 |… |j |… |b |
| |[pic] |[pic] |… |[pic] |… |[pic] |
| |i |pi1 |… |pij |… |pib |
| |[pic] |[pic] |… |[pic] |… |[pic] |
| |a |pa1 |… |paj |… |pab |
Then Ho: Variables A & B are independent is equivalent to:
Ho: pi1 = pi2 = … = pij = … = pib FOR EVERY ROW
Alternatively you could hypothesize Ho: p1j = p2j = … = pij = … = paj FOR EVERY COLUMN. It does not matter which version you use!
Example #2 Gender and Grad School Plans
Let’s determine if Gender and Grad School Plans are dependent or independent for STA 301 students.
0. pF = prop of Females going to Grad School & pM = prop of Males going to Grad School
1. Ho: Gender and Graduate School plans are independent ( pF = pM
2. HA: Gender and Graduate School plans are NOT independent. ( pF ( pM.
3. Set α = 0.05.
4. Calculate test statistic but also check assumptions!!!
| |Not Going to GSt |Going to GS |Total |
|Female |14 |35 |49 |
|Male |35 |39 |74 |
|Total |49 |74 |123 |
5.
6.
7.
THE TEST OF INDEPENDENCE ON SAS
OPTIONS LS=110 PS=60 PAGENO=1 NODATE;
TITLE 'CHI SQUARE HANDOUT EXAMPLE.SAS';
TITLE2 'STA301 GENDER AND GRAD SCHOOL';
PROC IMPORT REPLACE OUT=STA301
DATAFILE='C:\MyDocs\Class\STA 301\Surveys\Results\AllSTA301.xls';
PROC FREQ DATA=STA301;
TABLE GENDER*GRADSCHOOL/CHISQ EXPECTED NOPERCENT NOCOL;
RUN;
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
CHI SQUARE HANDOUT EXAMPLE.SAS 1
STA301 GENDER AND GRAD SCHOOL
The FREQ Procedure
Table of GENDER by GRADSCHOOL
GENDER(GENDER)
GRADSCHOOL(GRADSCHOOL)
Frequency‚
Expected ‚
Row Pct ‚NO ‚YES ‚ Total
ƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆ
FEMALE ‚ 14 ‚ 35 ‚ 49
‚ 19.52 ‚ 29.48 ‚
‚ 28.57 ‚ 71.43 ‚
ƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆ
MALE ‚ 35 ‚ 39 ‚ 74
‚ 29.48 ‚ 44.52 ‚
‚ 47.30 ‚ 52.70 ‚
ƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆ
Total 49 74 123
Frequency Missing = 2
Statistics for Table of GENDER by GRADSCHOOL
Statistic DF Value Prob
ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ
Chi-Square 1 4.3131 0.0378
Likelihood Ratio Chi-Square 1 4.3974 0.0360
Continuity Adj. Chi-Square 1 3.5672 0.0589
Mantel-Haenszel Chi-Square 1 4.2780 0.0386
Phi Coefficient -0.1873
Contingency Coefficient 0.1841
Cramer's V -0.1873
Fisher's Exact Test
ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ
Cell (1,1) Frequency (F) 14
Left-sided Pr = F 0.9888
Table Probability (P) 0.0176
Two-sided Pr Basic Statistics > 2 Proportions… option and the gender/computer question, we obtain:
Test and CI for Two Proportions
Sample X N Sample p
1 94 106 0.886792
2 43 60 0.716667
Difference = p (1) - p (2)
Estimate for difference: 0.170126
95% CI for difference: (0.0411347, 0.299117)
Test for difference = 0 (vs not = 0): Z = 2.77 P-Value = 0.006
We would conclude that the point estimate of the difference between the two proportions is 17.0%, with the interpretation that 17% more of the Females have computers compared to Males.
Our interval estimate indicates that this difference could be as small as 4.1% to as large as 29.9%.
We can conclude that at the minimum, 4% more of the Females have computers on campus compared to Males and at the maximum, this difference could be as large as 30%.
NOTES AND COMMENTS
1. One final comment is in order for the Chi-Square Test of Independence. As with every test one does (and this is also true of any confidence interval!), there are certain assumptions that must be met in order for the conclusions to be statistical valid. For every statistical inference, we require that we have a random sample and if we have more than one random sample, that the samples are independent of one another.
The Chi-Square Test of Independence is what is known as a “large sample” test. The results are valid if the sample sizes are large, but not in the “ > 30 “ sense that we have seen previously. Reliable statistical packages (and Minitab is one!) will inform you when your results are questionable. Not only will Minitab inform you, it will not provide any statistical results since the results are very questionable. For example, below is the Chi-Square Test of Independence results for GENDER and the Boxer/Brief question BOXERS, with questionable results.
Tabulated statistics: GENDER, BOXERS
Rows: GENDER Columns: BOXERS
NO YES All
FEMALE 105 1 106
99.06 0.94 100.00
104.08 1.92 106.00
MALE 58 2 60
96.67 3.33 100.00
58.92 1.08 60.00
All 163 3 166
98.19 1.81 100.00
163.00 3.00 166.00
Cell Contents: Count
% of Row
Expected count
Pearson Chi-Square = 1.233, DF = 1, P-Value = 0.267
Likelihood Ratio Chi-Square = 1.171, DF = 1, P-Value = 0.279
* NOTE * 2 cells with expected counts less than 5
Note that Minitab warns us that our results are questionable. If we inspect our results, we only had three students total out of the 166 surveyed who objected to such a question. If we consider the population of students who object to such a question, our sample size, three in this case, is way TOO small to make any informed decisions! Hence we should view our statistical results with caution or perhaps not even report them!
The moral of the story is that every statistical inference (test or CI) has assumptions. It is your responsibility to check and verify that these assumptions are met and to only present those results that are statistically valid.
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Each of these are “cells.”
This means to “sum” over the second subscript so that
ni( = n11 + n12 + … + n1b = [pic].
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