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7/27/2019 Week 7 - T-Tests & Z-Tests
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Sampling distributions
Way of estimating population based statistics
Thus far we have evaluated everything at level of a sample
Now have means from SAMPLES rather than individual observations
Central Limit Theorem
As N increases mean approaches true population
parameter
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Directional vs. Non-Directional
Remember:Directional hypothesis specifies > or and< the mean
One-tailed tests for an effect in 1 direction (i.e.,> or
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Sampling Distributions,
Standard Error, & df
Sampling Distribution = theoretical distribution of
a sample statistic (e.g., sample mean)
Standard Error = Standard deviation of the
sampling distribution
df = n-1
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Hypothesis Testing: contd Hypothesis testing Is there less than a 5% probability
that our sample statistic (e.g., sample mean) came from apopulation with
1) Identify DV/IV2) Determine Hx in symbols
3) What are the expectations under H0
- set up sampling dist.
4) Calculate obtained test statistic (z or t)
5) Conclusions- Statistical: Reject H0?
- Substantive: Explain result in laymans terms
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Z-Test
Conducting a Z-test:
We must already know and
Draw one sample (and perhaps provide some treatment or
manipulation)
Compare sample mean to to determine if there is a difference
Convert sample mean into z-test score isp< 0.05
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Z-Tests Research often shows that stress produces negative
reactions in the form of depression, anxiety, behaviorproblems, etc. But in a study of the families of cancerpatients, Compas and others (1994) observed that young
children do not report an unusual amount of symptoms ofanxiety or depression. In fact, they even look a little betterthan average. Is it really true that young childrensomehow escape the negative consequences of this kind
of family stressor? Can you think of alternativehypotheses that might explain the results?
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Z-Tests One of the commonly used measures of anxiety in
children is called the Childrens Manifest AnxietyScale (CMAS; Reynolds & Richmond, 1978). Nineitems from this scale form the Lie Scale which are
intended to identify children who are giving sociallydesirable responses rather than responding honestly.The mean on this scale for school-aged children is3.87 with a standard deviation of2.61, according toReynolds &Richmonds research. Compas et al. (1994)collected data from a child in each of36 families inwhich one parent had recently been diagnosed withcancer. Each child completed the CMAS, and their LieScale scores were computed. The mean for this groupof children was 4.39
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Z-Test Could it be that young children under stress have low
anxiety scores not because they have very little anxiety,but because the anxiety is masked by an attempt to givesocially appropriate answers?
What information do we have to use?
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Z-Test
4.39 3.87 = 1.19
2.61/ 36
St. Deviation
Sample Mean Pop. Mean ()
N
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Statistical Conclusion
Because the obtained z (1.19) is less than thecritical z (1.96), we fail to reject the null
hypothesis .
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Substantive Conclusion
On average, children with at least one familymember diagnosed with cancer had lie scores thatdid not differ significantly from those for children
who
s parents did not have cancer .
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Z-Tests
Important Notes:
We rarely know value, so z-tests not very practical
When is not known, use T- test
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One-Sample T-Test
Statistical procedure used to show the avg
difference between the sample mean and thepopulation
Know a priori, but NOT
How far sample mean ( ) is from population inS.E. units
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When to use t over z Z:
- 1 sample of data
- know AND
- rarely used (because we usually dont know )
T:
- 1 sample of data- know
- use SD from sample to estimate
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Mini-Golf Example There are 18 holes at the mini-golf palace. The average
person scores a 50, meaning it took them 2.78 strokes toget their ball in each hole. A researcher devised a strategybook that claims to help people improve their mini-golfgame. To test this, the researcher evaluates the scores of
five individuals after they read the strategy book andcompared them to the average person. Their scores arebelow:
40
4542
36
39
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Mini-Golf Example
One-Sample Statistics
N Mean Std. Deviation Std. Error Mean
Score 5 40.4000 3.36155 1.50333
One-Sample Test
Test Value = 50
t df Sig. (2-tailed)
Mean
Difference
95% Confidence Interval of theDifference
Lower Upper
Score -6.386 4 .003 -9.60000 -13.7739 -5.4261
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Critical T (t*)
Distribution of t-scores is different from typical
normal dist.
Greater spread = shorter and fatter curve
Greater probability in the tails and less in center
As df increase, t-curve approaches that of normaldistribution
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Statistical Conclusion
Because the obtained t (4)=-6.39 exceeds thecritical t(4)=2.776, we reject the null hypothesis in
favor of the research hypothesis
Because the obtainedp = .003 is less than .05, wereject the null hypothesis in favor of the research
hypothesis.
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Substantive Conclusion
On average, mini-golfers who read the strategybook performed significantly better than those
who had not.
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Weather Example
May is a fickle weather month in Colorado. A
meteorologist suspects that in Boulder the last 12Mays have been warmer than average. The
average May high historically has been 65.3degrees. Is he correct in thinking the temperaturehas been unusually warm?
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Weather ExampleAverage May highs since 1992.
75.376.259.168.788.282.570.572.875.168.380.062.8
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Weather Example
One-Sample Statistics
N Mean Std. Deviation Std. Error Mean
Temperature 12 73.2917 8.16494 2.35702
One-Sample Test
Test Value = 65.3
t df Sig. (2-tailed) Mean Difference
95% Confidence Interval of the
Difference
Lower Upper
Temperature 3.391 11 .006 7.99167 2.8039 13.1794
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Statistical Conclusion
Because the obtained t(11)=3.39 exceeds thecritical t(11)=2.20, we reject the null hypothesis in
favor of the alternative
Because the obtainedp = .006 is less than .05, wereject the null hypothesis in favor of the
alternative.
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Substantive Conclusion
On average, temperatures in May since 1992have been significantly higher than the historical
average for Colorado.