week 7 - t-tests & z-tests

7/27/2019 Week 7 - T-Tests & Z-Tests

1/25


2/25

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


3/25

Directional vs. Non-Directional

Remember:Directional hypothesis specifies > or and< the mean

One-tailed tests for an effect in 1 direction (i.e.,> or


4/25

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


5/25

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


6/25

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


7/25

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?


8/25

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


9/25

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?


10/25

Z-Test

4.39 3.87 = 1.19

2.61/ 36

St. Deviation

Sample Mean Pop. Mean ()

N


11/25

Statistical Conclusion

Because the obtained z (1.19) is less than thecritical z (1.96), we fail to reject the null

hypothesis .


12/25

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 .


13/25

Z-Tests

Important Notes:

We rarely know value, so z-tests not very practical

When is not known, use T- test


14/25

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


15/25

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


16/25

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


17/25

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


18/25

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


19/25


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.


20/25


On average, mini-golfers who read the strategybook performed significantly better than those

who had not.


21/25

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?


22/25

Weather ExampleAverage May highs since 1992.

75.376.259.168.788.282.570.572.875.168.380.062.8


23/25

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


24/25


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.


25/25


On average, temperatures in May since 1992have been significantly higher than the historical

average for Colorado.

week 7 - t-tests & z-tests

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