Lecture 6
Hypothesis Testing and One Population Tests
Howell Ch 8, 12
Statistical Inference
Two processes
I. Estimation of the value of a population parameter.
Question asked: What is the value of the population parameter?
Answer: A number, called a point estimate, or an interval, called a confidence interval. The
interval is one that has a prespecified probability (usually .95) of surrounding the population
parameter.
Example: What % of persons think we should bomb selected targets in Syria?
Answer might be: 38% with a 5% margin of error.
This is a combination of point estimate (the 38%) and a interval estimate (from 38-5 to 38+5 or from
33 to 43%). The 5% means that the interval has probability .95 of surrounding the true population
percentage. From this we could conclude that it’s quite likely that the population does NOT want
strikes.
II. Hypothesis Testing. Null Hypothesis Significance Testing (NHST)
A. With one population.
Deciding whether a particular population parameter (usually the mean) equals a value specified by
prior research or other considerations.
Example: A light bulb is advertised as having an “average lifetime” of 5000 hours.
Question: Is the mean of the population of lifetimes of bulbs produced by the manufacturer equal to
5000 or not?
B. With two populations.
Deciding whether corresponding parameters (usually means) of the two populations are equal or not.
Example: Statistics taught with lab vs. Statistics taught without lab.
Question: Is the mean amount learned by the population of students taught with a lab equal to the
mean amount learned by the population of students taught without lab?
C. Three populations.
Deciding whether corresponding parameters (usually means) of the three populations are equal or
not.
And on and on and on.
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Introduction to Hypothesis Testing
Suppose we have a single population, say the population of light bulbs mentioned above.
Suppose we want to determine whether or not the mean of the population equals 5000.
Two possibilities
0. The population mean equals 5000, i.e., it is not different from 5000.
1. The population mean does not equal 5000, i.e., it is different from 5000.
These possibilities are called hypotheses.
The first, the hypothesis of no difference is called the Null Hypothesis: H0.
The second, the hypothesis of a difference is called the Alternative Hypothesis: H1.
Our task is to decide which of the two hypotheses is true.
Two general approaches
1. The Bill Gates (Warren Buffet, Carlos Slim Helu’) approach.
Purchase ALL the light bulbs in the population. Measure the lifetime of each bulb. Compute the
mean. If the mean equals 5000, retain the null. If the mean does not equal 5000, reject the null.
Problem: Too many bulbs. Cost would be enormous. Might not even be possible to get all of them.
2. The Plan B approach.
Take a sample of light bulbs.
Compute the mean of the sample.
Apply the following, intuitively reasonable decision rule
If the value of the sample mean is “close” to 5000, decide that the null must be true.
If the value of the sample mean is “far” from 5000, decide that the null must be false.
But how close is “close”? How far is “far”?
What if the mean of the lifetimes of 25 bulbs were 4999.99?
What if the mean of the lifetimes of 25 bulbs were 1003.23?
What if the mean of the lifetimes of 25 bulbs were 4876.44?
Clearly we need some rules.
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The p-value.
Statisticians have decided to redefine “close” and “far” in terms of probabilities, specifically,
probabilities computed under the assumption that the null hypothesis is true.
They first compute the probability of an outcome as extreme as the sample outcome if the null
were true. That probability is called the p-value for the experiment.
The p-value is also called the probability of the obtained outcome due to chance alone.
The significance level.
Statisticians then choose a criterion representing a comfortably small probability. That value is
called the significance level of the statistical test. It is typically .05 or some value smaller than .05.
They then use the following rule:
Loosely . . .
If the p-value is small, then reject the null hypothesis
If the probability of the observed outcome due to chance alone is small, then it
must not be due to chance, so reject the null (chance) hypothesis.
If the p-value is large, then retain the null hypothesis.
If the probability of the observed outcome due to chance along is large, then it
must be due to chance, so retain the null (chance) hypothesis.
Formally:
If the p-value is less than or equal to the significance level, then reject the null.
If the p-value is larger than the significance level, then retain the null.
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The Steps in Hypothesis Testing
Is the Population mean = 5000 or not?
Step 1. From the problem description, state the null hypothesis required by the problem and the
alternative hypothesis.
Step 2. Choose a test statistic whose value will allow you to decide between the null and the
alternative hypothesis.
Step 3. Determine the sampling distribution of that test statistic under the assumption that the null
hypothesis is true. We need the sampling distribution in order to get the p-value.
Once you've determined the sampling distribution, you'll know what values of the test statistic would
be expected to occur if the null were true and what values would not be expected if the null were
true.
Step 4. Set the significance level of the test.
Step 5. Compute the value of the test statistic.
Step 6. Compute the p-value of the test statistic.
The p-value is the probability of a value as extreme as the obtained value if the null were true,
i.e., the probability of the obtained outcome due only to chance. In most cases, it will be printed
by SPSS. Unfortunately, it will be labeled "Sig." not p.
Step 7. Compare the p-value with the significance level and make your decision.
The key steps graphically . . .
Articulate Null
Hypothesis and
Alternative
Hypothesis
Reject
Null
Yes
Gather
Data
Compute
p-value
p<=
Significance
level?
Inference and One Population Tests - 4
No
Do not
reject Null
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Worked Out Example Start here on 9/29/15.
We'll start with the test of a hypothesis about the mean of a single population pretending that we
know the value of sigma. (Text Ch. 11).
I. A consumer organization decides to test the truth of a manufacturer’s claims concerning the
lifetimes of its “CF” – Compact Florescent bulbs. These bulbs are advertised to have an average
lifetime of 5000 hours. Note that the manufacturer is not claiming that any specific bulb will last
5000 hours but that the average of the population of lifetimes will be 5000.
Suppose that it was known that the standard deviation of lifetimes to burn-out in the population
is 250, based on several years of research with CF bulbs.
The magazine employees purchase 25 bulbs and burn them all until they fail. The mean of the
sample is not yet known.
Note that the observed mean is not shown above. This is done to emphasize that in real life, we set
up the hypothesis testing process before knowing the result of the experiment.
Step 1: Null hypothesis and Alternative hypothesis.
Hmm. We want to know whether the mean of the lifetime of the bulbs is equal to 5000. This
translates to . . .
Null Hypothesis:
The mean of the population equals 5000.
µ=5000.
Alternative Hypothesis:
The mean of the population does not equal 5000.
µ≠5000.
Step 2: Test Statistic: In this particular problem, we have a choice. We could use X-bar, the mean
of the sample as our test statistic. But statisticians have found that it’s simpler to use the Z-statistic
computed from X-bar. It turns out that it's easier to use the Z-statistic from the start, since we
ultimately have to compute it anyway.
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Step 3. Sampling distribution of test statistic if null is true.
For this step you have to pretend that the null hypothesis is true and then ask yourself, "If I took
1000's of samples of 25 scores in each sample from the population, what would be the distribution of
my test statistic be?" The answer is as follows
We will use Z=(X-bar-5000)/(250/5) = (X-bar-5000)/50 as the test statistic. Recall that 250 is the
population standard deviation and 25 is the sample size, so sigma/sqrt(N) = 250/sqrt(25)=250/5=50.
Name of Sampling
Distribution if the null hypothesis is true =
___Normal__
Mean of sampling distribution if null is true =
______0________
Standard Deviation of sampling distribution if null is true =
______1________
The distribution is the normal from the Central Limit Theorem.
The mean of the distribution of Z's is 0 from our study of Z-scores.
And the standard deviation of Z's is 1, again from our study of Z's.
Step 4: Significance Level
This one's easy. It's quite common to use .05 as the criterion for the probability of the observed
outcome. If that probability is less than or equal to .05, we reject. If that probability is larger than
.05, we retain the null. So let the significance level be .05.
Step 5: Computed value of test statistic
From the problem description we find that the mean of the sample is 4925.
So X-bar = 4925
If X-bar = 4925,
“Z” = (4925-5000) / 250/5
= (4925-5000)/50 = -75/50 = -1.5
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Step 6: p-value. In this case, compute the probability of a value as extreme or more extreme than
the obtained value if the null were true.
Distribution of Zs under assumption that the null is true
-2
-2
Z=
Tail
area =
p-value =
-1
-1
00
-1.5
22
11
1.5
.0668
.0668
.0668 + .0668 =.1336
This is the probability of getting a Z-score as extreme as 1.5 in either direction by chance alone.
Step 7: Conclusion
Since the p-value is larger than .0500, do not reject the null hypothesis. A Z as extreme as 1.5 could
have been obtained by chance alone.
We have no evidence that the population mean of lifetimes is different from 5000.
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Possible Outcomes of the HT Process
State of World
Null True µ=5000
Null False µ≠5000
Reject Null
Incorrect Rejection
Type I Error
Correct Rejection
Retain Null
Correct Retention
Incorrect Retention
Type II Error
Decision
Type I Error: Rejecting a true null hypothesis.
The Null is true, the population mean really is 5000 but was incorrectly rejected. How can that
happen? Because of a random accumulation of factors, our outcome was one which seemed
inconsistent with the null. So we rejected it - incorrectly. The pop mean of lifetimes was 5000, but
chance factors caused the sample mean to be very far from 5000, leading to rejection of the null.
Controlling P(Type I Error): The significance level determines the probability of this type of error.
This type of error will occur with probability given by the significance level - usually .05. That is, if
you always employ a significance level of .05, then 5% of those times when the null actually is true,
you will incorrectly reject it.
Type II error: Retaining a false null hypothesis.
The Null is false but was incorrectly retained. How could that happen? Because of a random
accumulation of factors, chance factors working backwards, our outcome was one which appeared
consistent with the null. So our decision was to not reject it – an incorrect decision. The
manufacturer was lying, but we incorrectly concluded that the advertising was true. A difference was
"missed".
We’ll discuss this type of error in the lecture on Power. The probability of this type of error should
be kept as small as is possible.
Controlling P(Type II error). We have less control over the size of the probability of a Type II error
than we do over the probability of a type I error. Sample size is easiest to manipulate. Larger
sample sizes lead to smaller P(Type II error). More on this in the section on Power.
Correct Retention: Retaining a true null hypothesis. Start here on 10/6/15.
The null is true. It should have been retained and it was. This is affirming the null. It is a correct
decision.
The manufacturer’s claim was correct, and we concluded that it was correct.
Controlling: Significance level.
Correct Rejection: Rejecting a false null hypothesis
The null is false. It should have been rejected, and it was. This is where we want to be. The null
was false. We "detected" the difference.
The manufacturer’s claim was not correct, and we detected that it was not correct.
The probability of correctly rejecting a false Null is called the Power of the statistical test. Power
equals 1 - P(Type II Error). More on power later.
Controlling: Same factors as those that affect P(Type II error). Sample size is most frequently
employed.
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Estimation
Using information from samples to guess the value of a population parameter or difference between
parameters.
Point Estimate
A single value which represents our single best estimate of the value of a population parameter.
Point estimate of Pop mean is sample mean. Point estimate of Population variance is SN-12.
Interval Estimate
An interval which has a prespecified probability of surrounding the unknown parameter.
Typically centered around the point estimate.
<---------Confidence Interval---------->
Lower limit
Upper limit
Point estimate
statistic
Statistics used as point estimates: Sample mean, Difference between two sample means, Pearson r
Estimation and Hypothesis Testing
A statistical law that relates estimation to hypothesis testing is the following:
If you have a 95% confidence interval for a population parameter,
1) you reject at significance = .05 the null hypothesis that the population mean equals ANY
value outside the confidence interval.
2) you retain at significance = .05, the null hypothesis that the population mean equals ANY
value Inside the confidence interval.
Suppose the confidence interval for the light bulb problem is 4827 – 5023.
Since this interval includes 5000, this means that we retain the null hypothesis that the population
mean = 5000.
On the other hand, if the null hypothesis was that Population mean = 4500 you would reject that
hypothesis.
Similarly if the null were that the population mean = 5500 you would reject.
Etc.
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Rationales for computing confidence intervals rather than doing hypothesis tests –
1. From the above, once you have a confidence interval, you can use it to test any hypothesis you
wish, so it’s more efficient to simply compute a confidence interval.
2. A confidence interval gives us a perspective on the location of the population mean.
A hypothesis test gives us only a “Yes” vs “No” answer.
But a confidence interval gives us that “Yes” vs “No” result along with some perspective.
Example. Both of the following hypothetical confidence intervals exclude 0, implying that the null
hypothesis that the population mean equals 0 would be rejected. But the one on the right gives
greater assurance that the population mean is not 0 than does the one on the left.
CI
--------------0-(------------)------------------Limits of CI: (.002, .270)
CI
---0-------------(------------)--------------(.123, .347)
The CI tells us how far the hypothesized population mean value is from our outcome.
If it is just outside the CI, that gives us the impression that our hypothesis was “almost” correct.
If it’s WAY outside the CI, gives us the impression that our hypothesis was “way” incorrect.
3 . If multiple hypothesis tests on the same phenomenon are conducted, some will reject and some
will retain the null. This leads to the appearance of inconclusiveness on the part of data analysts.
The report of multiple confidence intervals which will probably overlap gives a more coherent
picture of the phenomenon. Suppose there were 4 tests of the same hypothesis.
Experiment 1
Experiment 2
Experiment 3
Experiment 4
Hypothesis Tests
CIs
Reject
(.002, .270
Retain
(-.003, .265
Reject
(.005, .275
Retain
(-.004, .285)
CIs Graphically
----------0-(-----------)------------------------------(-0----------)----------------------------------0---(----------)------------------------------(-0---------------)---------------------
Looking at just the Reject – Retain – Reject – Retain outcomes of the hypothesis tests leaves you
feeling as if the researcher doesn’t know what’s going on.
But looking at the confidence intervals (Mike – pull aside the white square) shows that those
opposite conclusions might have been all identical if the data had been just slightly different from
one Experiment to the next.
Reporting confidence intervals gives us the opportunity to see how close different experimental
results were to each other.
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Example involving One Population Mean - value of sigma is known
A manufacturer of tires claims that average lifetime of its tires is 30,000 miles. A sample of 25 tires
is purchased and the tires are put on test cars that are driven until the tire tread reaches the minimum
legal depth. Based on the results of the sample, what can be concluded about the manufacturer's
claim? Suppose that years of experience with the tire indicated that the population standard
deviation is 900.
Step 1. From the problem description, state the null hypothesis required by the problem and the
alternative hypothesis.
Null:
Mean of the population of mileages of the tires is 30,000.
Alternative: Mean of the population of mileages is not 30,000.
Step 2. Choose a test statistic whose value will allow you to decide between the null and the
alternative hypothesis.
We’ll use the one sample Z statistic,
Z=
(X-bar – 30,000)
---------------------------Sigma / square root of N
The standard error of the mean is 900/sqrt(25) = 900/5 = 180.
Step 3. Determine the sampling distribution of that test statistic under the assumption that the null
hypothesis is true.
From our previous discussions, the distribution of the Z statistic, if the null is true, is the Normal
distribution with mean = 0 and standard deviation equal 1.
Step 4. Set the significance level of the test.
We’ll use .05.
Step 5. Compute the value of the test statistic.
Suppose that the mean of the sample of 25 tires was 29,640.
Since the sample mean is 29,640, Z = (29,640 – 30,000) / 180 = -360/180 = -2.00.
Step 6. Compute the p-value of the test statistic.
The p-value is the probability of a Z as extreme as -2.00
P(Z <= -2.00) = .0228. P(Z >= +2.00) = .0228.
P(Z as extreme as -2.00 = P(Z <= -2.00) + P(Z >= +2.00) = .0228 + .0228 = .0456.
Critical Z values In the precomputer days, data analysts determined the Z value whose p would be
just equal to .05. That value was called the critical Z. Any obtained Z larger than the critical Z in
absolute value would have a p less than or equal to .05.
The critical Z corresponding to alpha = .05 is 1.96. So any obtained Z larger than or equal to 1.96
will have a p less than or equal to .05.
Nowadays, those who do a lot of hypothesis testing may remember that the critical Z corresponding
to significance level is 1.96.
But for other tests, we’ll let our computer program compute the p-value directly.
Step 7. Compare the p-value with the significance level and make your decision.
Since p < .05, Reject the null hypothesis that Population mean = 30,000. Conclude that
population mean is less than 30,000.
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Confidence Interval on the Population mean Skip the details in 2015
1. Compute the point estimate. In this case, that's the sample mean, 29,640.
2. Decide on the level of confidence desired. How about 95%.
3. Use the following formulas to compute the upper and lower limits of the interval using the
following formula's
LL = X-bar + ZC1 * Standard error of X-bar
UL = X-bar + ZC2 * Standard error of X-bar
ZC1 and ZC2.are the values equal distance from the mean between which 95% of the scores fall.
(If this were a theoretically oriented course, we would show how ZC1 and ZC2 are dervied.)
In the case of normally distributed quantities, these values are -1.96 and +1.96.
So, for our problem, the 95% confidence interval would be
LL = 29,640 - 1.96*900/5
UL = 29,640 + 1.96 * 900/5
LL = 29,640 - 352.80
UL = 29,640 + 352.80
LL = 29,287.2
UL = 29,992.8
The probability is 95% that the interval 29,287.2 to 29,992.8 surrounds the population mean.
Graphically
30,000
|
------------------------------------(--------------------------)-----------------------------------------------------|
|
29, 287
29,993
Loosely, we can be 95% sure that the population mean falls in the interval 29,287 to 29,993.
And we can see from the graphical representation that the confidence interval did not include
30,000.
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Example involving One Population Mean - value of sigma not known
A manufacturer of tires claims that average lifetime of its tires is 30,000 miles. A sample of 25 tires
is purchased and the tires are put on test cars that are driven until the tire tread reaches the minimum
legal depth. Based on the results of the sample, what can be concluded about the manufacturer's
claim? Suppose that you don’t know the value of the population standard deviation. Suppose
that the mean of the sample of 25 tires was 29,640 with sample standard deviation equal to 900.
Step 1. From the problem description, state the null hypothesis required by the problem and the
alternative hypothesis.
Null:
Alternative:
Mean of the population of mileages of the tires is 30,000.
Mean of the population of mileages is not 30,000.
Step 2. Choose a test statistic whose value will allow you to decide between the null and the
alternative hypothesis.
We can’t use the Z statistic, because its formula requires that we know the value of the population
standard deviation.
A natural substitution for the population standard deviation is the sample standard deviation.
This would yield the formula,
Statistic =
(X-bar – 30,000)
--------------------------------Sample SD / square root of N
Step 3. Determine the sampling distribution of that test statistic under the assumption that the null
hypothesis is true.
The statistic above is NOT a Z, since S, rather than sigma has been used in the denominator.
So the sampling distribution is not the Normal Distribution, as it would be if the statistic were Z.
The actual formula for the sampling distribution was discovered by William Gossett, in the early
1900’s.
He called the statistic the t statistic and called the distribution the T distribution.
The T distribution has mean 0.
Its standard deviation is larger than 1, and depends on a quantity called degrees of freedom.
In the one-sample case, df = N – 1 = 24 for our example.
The shape of the T distribution is similar to the Normal.
Step 4. Set the significance level of the test.
We’ll use .05.
Step 5. Compute the value of the test statistic.
Since the sample mean is 29,640,
t = (29,640 – 30,000) / (900/5) = 360/180 = -2.00.
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Step 6. Compute the p-value of the test statistic.
The p-value is the probability of a t as extreme as -2.00.
The formula for computing the p-value for a t is quite complicated. Luckily, we have computers.
Our computer program will compute the value of t and its p-value automatically. That p is .057.
Or we could look up the critical t for df = 24 and alpha = .05. That critical t value, from the text, is
2.064.
If the obtained t exceeds the critical t in absolute value that would mean that p would be less than or
equal to .05 and we could reject the null.
Step 7. Compare the p-value with the significance level and make your decision.
Since p > .05, do not reject the null hypothesis that Population mean = 30,000. We will act as if
the population mean is equal to 30,000.
So, we rejected the null hypothesis when using the Z statistic.
But we did not reject the null when using the t.
This illustrates the fact that the Z statistic is slightly more powerful than the t.
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Confidence Interval on the Population mean
1. Compute the point estimate. In this case, that's the sample mean, 29,640.
2. Decide on the level of confidence desired. How about 95%.
3. Compute the upper and lower limits of the interval using the following formula's
LL = X-bar + tC1 * Standard error
UL = X-bar + tC2 * Standard error
(Again, if this course were theoretically oriented, we would derive tC1 and tC2 here.)
tC1 and tC2.are the t values equal distance from the mean between which 95% of the scores fall.
In this case, these values are the two-tailed critical values that would be used if alpha were .05.
From Table B, Appendix C, p. 465, critical t when df=24 is 2.064. You won’t be tested over this.
So, for our problem, the 95% confidence interval would be
LL = 29,640 – 2.064*900/5
UL = 29,640 + 2.064 * 900/5
LL = 29,268.5
UL = 30,011.5
The probability is 95% that the interval 29,268.5 to 30,011.5 surrounds the population mean.
Graphically
30,000
|
------------------------------------(------------------------------)------------------------------------------------|
|
29, 287
30,011.5
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Comparing the Z-test and the t-test for the same data
Z-test results
t-test results
H0: Pop mean = 30000
H0: Pop mean = 30000
Pop SD = 900
Sample SD = 900
N = 25
N = 25
Sample mean = 29,640
Sample mean = 29,640
Z=
(29640 – 30000)/(900/5)
t = (29640 – 30000)/(900/5)
2.00
p=
2.00
.0456
p=
.0570
(----------------------------------------------------)
Z CI = ---------------------------|---------------------------|-------------------------|-------------------------29,287
29640
29,993
(--------------------------------------------------------------)
T CI = ----------------------|--------------------------------|------------------------------|------------------29,268
29640
30011
30,000
The bottom line is that the Z is a more powerful test statistic than the t. But the Z is higher
maintenance – it requires that you know the value of the population standard deviation in order to
achieve that power.
A tractor is a more powerful tool for plowing than a horse. But you
must have gasoline to take advantage of that power.
The t is more forgiving, but at the price of not giving you the precision of the Z.
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One Sample t – SPSS Example
A manufacturer of tires claims that average lifetime of its tires is 30,000 miles. A sample of 25 tires
is purchased and the tires are put on test cars that are driven until the tire tread reaches the minimum
legal depth. Based on the results of the sample, what can be concluded about the manufacturer's
claim? The mileage values for the tires are below.
id
1
2
3
4
5
6
mileage
29050
29837
29622
29546
28197
28277
7
8
9
10
11
12
29585
31344
29702
29585
29564
28824
13
14
15
16
17
18
31158
29084
29467
28261
28496
30784
Inference and One Population Tests - 17
19
20
21
22
23
24
25
2/9/2016
30534
30397
29883
29077
31239
29630
29857
Select the variable to
be analyzed and
move it to the “Test
Variable(s)” field.
Very Important. Put
the hypothesized
value of the
population mean in
the “Test Value”
field.
Choosing the confidence interval level.
Click on the
[Options] button to
open the Confidence
Interval dialog box.
Hypothesis Testing and One Population Tests - 18
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The Output
T-Test
Test value
entered
above.
One-Sam ple Stati stics
N
mil eage
25
Me an
Std . Deviatio n Std . Erro r Me an
296 40.0 0
899 .908
179 .982
One-Sam ple Test
Te st Val ue = 3000 0
95% Co nfide nce In terva l of
the Diffe rence
mil eage
t
-2.0 00
Value of t
df
24
Sig . (2-t ailed ) Me an Differe nce
.05 7
-36 0.000
The twotailed p-value.
Lower
-73 1.46
Up per
11. 46
Standardized Confidence
interval.
Add test value to each limit
to get it in terms of the
problem’s values.
Correct limit of confidence are
LL = -731.46 + 30,0000 =
29,268.54
UL = 11.46 + 30,000 =
30,011.46
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Hypotheses about the Population Correlation Coefficient, r
Biderman and Nguyen (2004) conducted research in which participants were given the Big 5
personality test twice, once under instructions to respond honestly and then again under instructions
to “fake good”. We computed a score on each of the Big 5 dimensions under the fake condition and
a score on each of the dimensions under the honest conditions. We subtracted the honest condition
scores from the faking condition scores and then computed the mean of those 5 differences. That
average was the respondents “Faking ability” score. A positive value means that the person was
able to fake good, responding with higher scores in the faking condition than in the honest condition.
A score of 0 meant that the respondent score no higher on the personality dimensions in the faking
condition than in the honest condition. And a negative score meant that the respondent actually
scored lower on the personality dimensions in the faking condition than in the honest condition.
We gave each respondent the Wonderlic, a common measure of cognitive ability. The distribution
of faking ability scores and scores on the Wonderlic are as follows . .
30
25
Frequency
20
15
10
5
0
-2.0
-1.0
0.0
1.0
2.0
Mean = 0.512
Std. Dev. = 0.7247
3.0 N = 203
Ability to fake the Big 5 defined as the mean
of the 5 F-H differences
3.0
2.0
1.0
Faking
Ability
wonderlic test score
40
30
Frequency
Ability to fake the Big 5 defined as the mean of the 5 F-H
differences
Ability to fake the Big 5 defined as the mean of the 5 F-H
differences
20
10
0.0
-1.0
R Sq Linear = 0.094
-2.0
Mean = 24.61
Std. Dev. = 6.753
N = 203
0
0
10
20
30
wonderlic test score
40
50
10
20
30
40
wonderlic test score
We then correlated the FAKABIL and WONDLIC scores. The scatterplot is above.
The question is:
In the population, is there a correlation between cognitive ability and faking ability?
Hypothesis Testing and One Population Tests - 20
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50
Testing Hypotheses on Population r
Null Hypothesis:
The population correlation coefficient is 0
Alternative Hypothesis:
The population correlation coefficient is not 0.
Test statistics: (There are two possibilities)
1) A t statistic
Sample r – 0
t = ---------------------1 – Sample r2
------------N - 2
Critical t values would be obtained from tables of critical t.
2) r itself.
Critical r values would be obtained from tables of critical r.
SPSS Analysis
Analyze -> Correlate -> Bivariate
r
Correla tions
gscore
fakabil Abilit y to
fake the Big 5
de fined as th e
gscore wond erlic
me an o f the 5
test score
F-H diffe rences
wond erlic test score Pe arson Correlatio n
1
.30 7**
Sig . (2-t ailed )
N
fakabil Abilit y to f ake th e
Big 5 de fined as th e me an
of the 5 F-H d iffere nces
.
.00 0
20 3
20 3
Pe arson Correlatio n
.30 7**
Sig . (2-t ailed )
.00 0
.
20 3
20 3
N
p-value
N
1
**. Co rrelat ion is sign ifican t at th e 0. 01 le vel (2 -taile d).
Conclusion: Our data suggest that in the population, the ability to fake a Big 5 personality test is
positively related to cognitive ability.
Confidence intervals for the Population r
Minium, et. al.’s Figure 16.7 on p. 302 is an excellent table. (I’ll copy and print it for you.)
Formulas required to determine the LLs and ULs are complicated.
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