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Hypothesis Testing and Chi
Square
STAT Chap 13
PSYC 201 – Psychological Research I
Inferential Statistics are used to make
three kinds of inferences
Point estimates A statistic (e.g., the mean) computed from a single sample of subjects is used to estimate the mean of the population of subjects Interval estimates or Confidence Intervals
An interval, with limits at either end, with a specified probability (usually 95%) of including the parameter being estimated (e.g., the population mean)
Hypothesis testing or Statistical Significance A process whereby you decide whether the differences or relationships observed in the data are due to chance fluctuations or whether they are sufficiently large enough to consider them “significant” or not due to chance
We’re ready for hypothesis testing!
We’re interested, right now, in going over the
general strategy of inferential statistics with
regards to hypothesis testing
And, we’re going to apply it to the chi square
test used to analyze categorical data
Inferential statistics are based on probability
statements
Probability statements are statistical
statements that
Describe the data
Predict what we don’t know, on the basis of
current data
So: inferential statistics are all about making inferences about the population based on sample data
Therefore: it is very important to acquire a sample of subjects or observations that truly reflects the population E.g: Population = all fourth grade children in the U.S. Sample = 100 fourth grade children from all elementary schools in Montgomery County
Sampling
How you select the participants in the sample
is extremely important:
In order to generalize from sample to population, the sample must be representative of the population
If the sample has not been generated randomly (each possible sample of that size has an equal chance of being selected), then the laws of probability won’t apply as well to the sample
Therefore: your statistical conclusions will be
faulty
Step 2 (Determine H 0 and H 1 )
Create the Null and Alternative hypotheses Null = H 0 = no effect or no relationship present Alternative = H 1 or H (^) a = effect or relationship present
The Null is always the opposite of what the researcher wishes to demonstrate is true (because one cannot prove something true, but one can show something to be false…)
Standard Operating Procedure (SOP) in Psychology is to use a non-directional hypothesis
Psychologists generally use non-directional
hypotheses (referred to as a two-tailed test), so that’s
what we do in this class, always.
Non-directional hypothesis examples:
H 0 : χ^2 = 0 H 1 : χ^2 ≠ 0
H 0 : μ 1 = μ 2 H 1 : μ 1 ≠ μ 2
Directional hypothesis examples:
H 0 : μ 1 = μ 2 or H 0 : μ 1 = μ 2 H 1 : μ 1 > μ 2 H 1 : μ 1 < μ 2
Step 3 (Determine the critical value)
Next, determine the cutoff point at which the null hypothesis (H 0 ) will be rejected, which is the criteria for significance
SOP for Psychologists is to use a cutoff of 5%, otherwise stated as an alpha of .05 (α = .05) If a result is significant at the .05 level, then the effect could have occurred by chance only 5 times or less out of 100 (probably)
At this point, you look up the critical value for your test statistic in the Appendix for your particular research problem (e.g., chi square)
Step 3 (continued)
Determine the rejection region using α =.
You will usually need the degrees of freedom ( df ) to obtain the critical value for the test statistic Then check in the appropriate Appendix in your STAT book (page 296)
If you’re doing a Goodness of Fit test: df = k – 1 k = number of categories df = 4 – 1 = 3 χ^2 critical = 7.
If you’re doing a Test of Independence: df = (r-1)(c-1) r = number of rows & c = number of columns
The Hypothesis Test:
When requested to set up the Hypothesis Test,
the following example is what is expected:
H 0 : χ^2 = 0 α =. H 1 : χ^2 ≠ 0 df = 4 – 1 = 3 χ^2 critical = 7.
The critical value for your chi square problem is the value of χ^2 that you have to obtain, or bigger, when you do your calculations in order to reject the null hypothesis
Step 4 (Data analysis)
Once the hypothesis test has been set up, then you check the data, enter it into the computer (as needed), summarize it, and compute inferential statistics to determine the obtained value of the test statistic
If done on the computer using SPSS, the program will give you the exact probability of obtaining that value of the test statistic for that number of subjects, assuming that the null hypothesis is true It is called Sig., short for significance level
One-sample chi square test: Goodness
of Fit
One category: season, with 4 levels
The rationale is that in any set of occurrences,
you can easily compute what you would expect
by chance:
Simply divide the total number of occurrences by
the number of classes or categories or levels
Therefore, 28 + 33 + 16 + 51 = 128
Chance?
If season makes no difference, or if the only thing operating is chance, then you would expect 32 clients each season (the null hypothesis)
How different are the observed numbers from the expected numbers?
Compute the chi square value and compare it to the critical value of chi square determined earlier to make the decision
Χ^2 = Σ
(O – E) 2
E
Χ^2 is the chi-square value
Computing chi square
Σ is the summation sign
O is the observed frequency
E is the expected frequency
Plug each O and E value into the formula (for each
cell), square, divide by E, and then sum the 4 values
up:
Winter Spring Summer Fall E O X^2 E O X^2 E O X^2 E O X^2 28 33 16 51
32 0.5 32 0.031 32 8 32 11.
Χ^2 = 19.
Step 5 (Make a decision)
Make the decision to reject or retain the null
hypothesis
You compare your criterion for significance,
the cutoff point or rejection region, that you
determined in Step 3, to the value and/or
probability that you obtained when you made
the calculations on your sample of data
Step 5 (continued)
Two ways to make the decision:
χ^2 obtained = 19.81 and χ^2 critical = 7.
Then |19.81| > |7.81|
Therefore, you reject the null hypothesis; and
conclude that it is significant
OR, if you did it on the computer, for example,:
sig. = .01 (on computer printout) and α =.
.01 < .05 therefore, reject the null; it is
significant
Outcomes of Decision Making
Type II Error p = β
Correct Decision p = 1 - α
Do not reject Null
Correct Decision p = 1 - β = power
Type I Error p = α
Reject Null
True state of affairsÆ Null is True Null is False
Your decision ↓
Psychologists generally consider the Type I
error more serious (you assume that there is
an effect in the population when in fact there
is not)
How do you decrease the risk of a Type I
error? We’ll get into this a bit later when we
talk about Statistical Power
Step 7 (Write the results in APA-style)
When you write up the results in an APA-style
paper you must include the following
information for each analysis conducted:
Describe the data analyzed Describe the statistical test used State whether or not the results were significant Report the STAT statement ( Χ^2 (1, N = 387) = 1.37, p = .242) Report the descriptive statistics Provide a conclusion
Chi square Test of Independence
When there are 2 variables, you set up a
contingency table
Example: In a study on the phenomenon of
“blaming the victim” in prosecutions for rape,
two variables were examined: 1) whether
the defense alleged that the victim was
partially at fault (Low vs. High fault), and 2)
whether or not the defendant was found guilty
or not. (Based somewhat on Pugh, 1983).
The Hypothesis Test for our Test of
Independence example:
H 0 : χ^2 = 0 α =.
H 1 : χ^2 ≠ 0
df = (r-1)(c-1) = (2-1)(2-1) = 1 x 1 = 1
df = 1 χ^2 critical = 3.
The number of observations in each
cell are presented below
Column 1 Column 2 Guilty Not Guilty Row E O X^ 2 E O X^ 2 Totals Low 153 24 177 Row 1 Fault
Row 2 High Fault Column 258 100 Totals
Make the decision
χ^2 obtained = 35.93 and χ^2 critical = 3.
Then | 35.93 | > |3.84|
Therefore, you reject the null hypothesis; and
conclude that it is significant
Draw, and write, conclusions:
The number of people that judged the defendant guilty on a charge of rape as a function of the alleged level of fault attributed to the victim was analyzed using a chi square Test of Independence. There was a significant effect such that the guilty verdict was not independent of the level of fault attributed to the victim, Χ^2 (1, N = 358) = 35.93, p < .05. If the victim was portrayed as high in fault the defendant was almost as likely to be found not guilty (21.23%) as guilty (29.33%). Whereas, if the victim was seen as being low in fault, then the defendant was much more often found to be guilty (42.74%) as opposed to not guilty (6.7%).
