Stats 95

Z-Scores & Percentiles

• Each z-score is associated with a percentile.– Z-scores tell us the percentile of a particular score– Can tell us % of pop. above or below a score, and

the % of pop. between the score and the mean and the tail.

Transforming Z-score into Percentiles

• Draw a Graph!…did I mention you need to draw a graph? Yeah, draw a graph.

a) DRAW A GRAPH!

b) Calculate z-score

c) Estimate the percentile of the z-score using probability distribution

d) Use z-score chart to transform z-score into percentile

e) Use graph to make sure answer makes sense

Transforming z-Scores into Percentiles• Use a chart like this in Appendix A of your text (Yes, you need the textbook) to find the percentile of you z-score.

• This table gives the distance between the mean (zero) and the z-score.

• To calculate cumulative percentile :

•Of positive z-score 50 + (z)

•Of negative z-score 50+ (-z)

Example: Height

• Jessica has a height of 66.41 inches tall (5’6’’)• The mean of the population of height for girls is 63.80• The standard deviation for the population height fir

girls is 2.66

• According to z-score table, the percentile associated with z = .98 is 33.65%

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Height Example:Did I mention? DRAW A GRAPH!!

• Jessica’s z-score for her height is .98, associated percentage of 33.65%.

• This means – there is 33.65% of the population is between the

mean and Jessica’s score.– There is a 33.65% chance of Jess being taller than the

average by this amount BY CHANCE ALONE

• Mean = 50th percentile, therefore to find the Jessica’s percentile = 33.65+50 = 83.65%.

• 84% of the population of girls is shorter than Jessica, and there is a 100-84% = 16% chance of someone being this tall by CHANCE ALONE.

Hypothesis Testing

• Identify Population and comparison group, state assumptions

• Define the Null Hypothesis

• Define the Research Hypothesis or Alternative hypothesis– Define the Research and Control Group

– Define the Dependent and Independent Variables

• State relevant characteristics of comparison distribution

• Determine critical cutoff values

• Calculate statistic

• Reject or Fail to Reject the NULL Hypothesis

Statistical Significance

• A finding is statistically significant if the data differ from what we would expect from chance alone, if there were, in fact, no actual difference.

• They may not be significant in the sense of big, important differences, but they occurred with a probability below the critical cutoff value, usually a z-score or p < .05

Assumptions for Parametric Tests

• Dependent variable is a scale variable interval or ratio– If the dependent variable is ordinal or nominal, it is a non-

parametric test

• Participants are randomly selected– If there is no randomization, it is a non-parametric test (and

likely a flawed experiment, or one limited in generalization)

• The shape of the population of interest is approximately normal– If the shape is not normal, it is a non-parametric test

Assumptions for Parametric Tests

Hypothesis Testing with z-Scores: Example

• Ursuline College, 97 students took Major Field Test in Psychology (MFTP). Is the score of this group of students statistically different* from the total population of students who took the test?

• *is the probability of the difference greater than what we would expect to happen by chance alone. (i.e., how big is their horseshoe?)

Hypothesis Testing with z-Scores Step 1 Populations, Distribution, Assumptions

• Identify Population, distributions and assumptions– All students at Ursuline who took

the exam

– All students nationally who took exam

• Distribution– Comparison sample is not an

individual but a group mean for all the students at Ursuline who took exam therefore comparison distribution is a sample mean to a population.

• Assumptionso Scale is continuous interval

(test scores)o Random selection

unknown, so we are limited in generalizing

o Shape should be normal, sample size of 97 Ursuline students, substantially larger than recommended 30

Hypothesis Testing with z-Scores: Step 2: State Null and research Hypothesis

Population Parameters

• The Null Hypothesis: Ursuline

Students are not better from rest of the nation.

• Research Hypothesis, U. students

are better than population

• One-tailed Test (Rare!)

Non-directional

• The Null Hypothesis: Ursuline

Students are no different from rest of the nation.

• Research Hypothesis, U. students

are Different (better or worse) than population

• Two-tailed test

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Hypothesis Testing with z-Scores: Step 3: Determine Characteristics (Parameters) of the Distributions

• Comparing Sample Mean to Population

– Null H0 and Research H1

– Because we have a sample mean, we must use the Standard Error instead of the Standard Deviation of the Population

– We have Standard Deviation of Population, from which we calculate the Standard Error of Sample Means

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– The average of Ursuline class was 156.11. The average of all other College mean samples from other college classes is 156.5.

– We are saying Null H. assumes the mean of this class is no different from the combined mean of all classes.

Hypothesis Testing with z-ScoresStep 4: Determine Critical Values (Set how big a Horseshoe You Will Willingly Accept)

• Convention is to set cutoffs at probability less than 5% – p < .05

– Two-tailed Test, means it is 2.5% at both Tails

– If it was a One-tailed Test, 5% at one Tail

• Draw a Graph

• We know 50% of the curve falls between the mean and each end. We know that 2.5% falls between the critical z statistic and the rest of the tail. Subtract 50%-2.5 = 47.5% . Look in chart, 47.5% is associated with the z statistics of 1.96

Hypothesis Testing with z-Scores:Step 5: Calculate Z Statistic

• Draw a Graph!

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Hypothesis Testing with z-Scores:Step 6: DECIDE Already!

• Decision is either to Reject the Null Hypothesis or to Fail to Reject the Null Hypothesis

• With a Z statistic of -.26, there was a approx. 40% of getting this score by chance alone, p = .3976.

• Compare to the cutoffs– z statistics: -.26 is not more

extreme than -1.96

– p: .3976 >.05

• We fail to reject the Null H.

• The probability of Ursuline producing the score of 156.11 is greater than .05– p >.05

– The Ursaline mean was not smaller, than we would expect from chance alone, than the population mean.

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• What problems did Fisher have to tackle in the designing the experiment that could determine if the Lady could distinguish between milk-then-tea and tea-then-milk cups of tea? She could guess randomly with a 50% probability of being correct. She could also make an error is spite of being able to distinguish, getting 9 our of 10. How many cups to present, in what order, how much to instruct the lady, and determine beforehand the likelihood of possible outcomes.

 What was the nature of research like before experimental methods became established? They were idiosyncratic to each scientist, lesser scientists would produce vast amounts of data would be accumulated but would not advance knowledge, such as in the inconclusive attempts to measure the speed of light. They described their conclusion and selectively presented data that demonstrated or was indicative of the point they wanted to make, without clear procedures to replicate and not with full transparency.

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• In examining the fertility index and rainfall records at Rothamsted, what two main errors did he conclude about the indexes and how did he express results about the rainfall? He examined the competing indexes, and showed when reduced to their elemental algebra, they were all versions of the same formula. And he found that the amount of rainfall was greater than the type of fertilizer used, that the effects for fertilizer and rainfall had been confounded.

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• What did Fisher conclude scientists needed to start with to do an experiment, and what would a useful experiment do? Scientists need to start with a mathematical model of the outcome of the potential experiment, which is a set of equations in which some of the symbols stand for numbers that will be collected as data, and other symbols stand for the overall outcomes of the experiment. Then a useful experiment allows for estimation of those outcomes.