Presentation is loading. Please wait.

Presentation is loading. Please wait.

1 Why do we need statistics? A.To confuse students B.To torture students C.To put the fear of the almighty in them D.To ruin their GPA, so that they don’t.

Similar presentations


Presentation on theme: "1 Why do we need statistics? A.To confuse students B.To torture students C.To put the fear of the almighty in them D.To ruin their GPA, so that they don’t."— Presentation transcript:

1 1 Why do we need statistics? A.To confuse students B.To torture students C.To put the fear of the almighty in them D.To ruin their GPA, so that they don’t get into grad school, have to buss tables and move back in with parents E.All of the above F.All of the above (and other tragic outcomes)

2 2 A positive optimistic view…  It is a tool that could help you succeed and move out of your parents house  There is nothing to fear but fear itself  You need a passing grade of C  Can help to get into grad school  It’s important to understand so you don’t get scammed…

3 3 The Caveat: Remember…  There are lies  There are d#$m (darn) lies and then  There are statistics  Magic

4 4 Statistics  The science of collecting, displaying and analyzing data  Based on quantitative measurements of samples  Allow us to objectively evaluate data  Descriptive  Inferential

5 5 Defining variability  Amount of change or fluctuation  Some variability is expected  Is the observed variability due to the usual variability among subjects from the population?  Or is the observed variability greater than the usual variability

6 6 Frequency Distribution Dependent variable Frequency (# of Subjects) From population Sample 1 Sample 2 Sample 3 0highest score Highest

7 7 Frequency Distribution Dependent variable Frequency (# of Subjects) Untreated groups of an experiment Control Experimental

8 8 Frequency Distribution Dependent variable Frequency (# of Subjects) Treated Groups ControlExperimental

9 Beginning steps of an Experiment  Sample from population  Hypothesis  Define variables  Assign subjects to conditions  Measure performance  Calculate means  Calculate variability

10 10 Heading Error: Calculating Variance Deviation from the mean for each subject (X i – X) (X i – X) 2 Square the deviation from the mean for each subject Add the squared deviations together (X i – X) 2 Sex SUBJE CT$ HEADGdeg Femalerat1-4.44.4-4.6821.86 Femalerat311111.933.71 Femalerat52.32.3-6.7845.90 Femalerat78.58.5-0.580.33 Femalerat96.96.9-2.184.73 Femalerat11-10.810.81.732.98 Femalerat13-10.910.91.833.33 Femalerat1517.817.88.7376.13 Malerat229.629.66.8647.09 Malerat4-18.518.5-4.2417.96 Malerat614.514.5-8.2467.86 Malerat858.258.235.461257.59 Malerat10-18.718.7-4.0416.30 Malerat12-17.317.3-5.4429.57 Malerat14-14.814.8-7.9463.00 Malerat1610.310.3-12.44154.69 Female = 158.96 Male = 1654.06

11 11 Heading Error: Calculating Variance Sex SUBJE CT$ HEADGdeg Femalerat1-4.44.4-4.6821.86 Femalerat311111.933.71 Femalerat52.32.3-6.7845.90 Femalerat78.58.5-0.580.33 Femalerat96.96.9-2.184.73 Femalerat11-10.810.81.732.98 Femalerat13-10.910.91.833.33 Femalerat1517.817.88.7376.13 Malerat229.629.66.8647.09 Malerat4-18.518.5-4.2417.96 Malerat614.514.5-8.2467.86 Malerat858.258.235.461257.59 Malerat10-18.718.7-4.0416.30 Malerat12-17.317.3-5.4429.57 Malerat14-14.814.8-7.9463.00 Malerat1610.310.3-12.44154.69 Compute the Variance (X i – X) 2 s2 =s2 =s2 =s2 = n - 1 s 2 Female = 22.71 s 2 Male = 236.29

12 12 Heading Error: Calculating standard deviation Sex SUBJE CT$ HEADGdeg Femalerat1-4.44.4-4.6821.86 Femalerat311111.933.71 Femalerat52.32.3-6.7845.90 Femalerat78.58.5-0.580.33 Femalerat96.96.9-2.184.73 Femalerat11-10.810.81.732.98 Femalerat13-10.910.91.833.33 Femalerat1517.817.88.7376.13 Malerat229.629.66.8647.09 Malerat4-18.518.5-4.2417.96 Malerat614.514.5-8.2467.86 Malerat858.258.235.461257.59 Malerat10-18.718.7-4.0416.30 Malerat12-17.317.3-5.4429.57 Malerat14-14.814.8-7.9463.00 Malerat1610.310.3-12.44154.69 Standard deviation s or SD = s 2 s Female = 4.77 s Male = 15.37

13 13 Heading Error: Calculating standard error of the Mean Sex SUBJE CT$ HEADGdeg Femalerat1-4.44.4-4.6821.86 Femalerat311111.933.71 Femalerat52.32.3-6.7845.90 Femalerat78.58.5-0.580.33 Femalerat96.96.9-2.184.73 Femalerat11-10.810.81.732.98 Femalerat13-10.910.91.833.33 Femalerat1517.817.88.7376.13 Malerat229.629.66.8647.09 Malerat4-18.518.5-4.2417.96 Malerat614.514.5-8.2467.86 Malerat858.258.235.461257.59 Malerat10-18.718.7-4.0416.30 Malerat12-17.317.3-5.4429.57 Malerat14-14.814.8-7.9463.00 Malerat1610.310.3-12.44154.69 Standard error of the Mean SEM = SEM Female = 1.68 SEM Male = 5.43 SD nnnn

14 14 Heading Error: Group Means with SEM

15 15 Heading Error: Group Means with 95% Confidence Interval Confidence intervals (CI) represent a range of values above and below our sample mean that is likely to contain the population mean; i.e., the true mean of the population is likely (we’re 95% confident) to fall somewhere within the CI range. CI = X± t crit SD nnnn ()

16 16 Heading Error: Group Means with SEM s 2 variance (s 2 )- average squared deviation of scores from their mean standard deviation (SD)- average deviation of scores about the mean standard error of the mean (SEM)- dispersion of the distribution of sample means SEM = SD nnnn SD = s 2 (X i – X) 2 s2 =s2 =s2 =s2 = n - 1

17 17 Choosing a significance level Significance level - A criterion for deciding whether to reject the null hypothesis or not. What is the convention? p <.05 ( level)What is the convention? p <.05 ( level) A stricter criterion may be required if the risk of making a wrong decision (a Type I error) is greater than usual. p <.01 or p <.001.A stricter criterion may be required if the risk of making a wrong decision (a Type I error) is greater than usual. p <.01 or p <.001. But there is a trade off in using a stricter criterion.But there is a trade off in using a stricter criterion.

18 18 Choosing a significance level Type II error – Failure to reject the null hypothesis when it is really false (). Concluding that the difference is due to chance variation when it is really due to the independent variableConcluding that the difference is due to chance variation when it is really due to the independent variable Power of the statistical test (1 - )Power of the statistical test (1 - )

19 19 RealityCheck Decision based on Statisical Results Fail to reject H o Reject H o H o is true Correct p = 1 -  Type I error p =  H o is False Type II error P =  Correct p = 1 -  Summary Chart

20 20 Time estimation experiment Time will go faster for people having fun than for those not having fun. Two group design: Fun - views cartoons with the captions for 10 min. No Fun – views cartons without captions for 10 min. H o = the time estimates of the two groups will be the same. H 1 = the fun group will have shorter estimates than the control group. Table 13-2 possible errors in the time estimation experiment (p.381, 6 th ed.) We conclude that there was no difference in the time estimates made by the “fun” and “no fun” groups even though the treatments did produce an effect. We conclude that there was a difference in the time estimates made by the “fun” and “no fun” groups even though the treatments produced little or no effect at all. What type of errors were made in the two descriptions? Type I Type II Type I Type II Type III – failure to accurately identify a type 1 or 2 error Note - The error has been corrected in the 7 th ed., p. 390. Type 1 = Reporting an effect that doesn’t really exist Type 2 = Missing an effect that does really exist

21 21 Questions to ask when selecting a test statistic Table 14-1 The parameters of data analysis ___________________________________________________ 1.How many independent variables are there? 2.How many treatment conditions are there? 3.Is the experiment run between or within subjects? 4.Are the subjects matched? 5.What is the level of measurement of the dependent variable? ___________________________________________________

22 22 Answers based on the water maze study Table 13-1 The parameters of data analysis ___________________________________________________ 1.How many independent variables are there? one 2.How many treatment conditions are there? one 3.Is the experiment run between or within subjects? between 4.Are the subjects matched? no 5.What is the level of measurement of the dependent variable? ratio ___________________________________________________

23 23 Levels of Measurement Ratio – a measure of magnitude having equal intervals between values and having an absolute zero point. Interval – same as ratio except that there is no true zero point. Ordinal – a measure of magnitude in the form of ranks (not sure of equal intervals and no absolute zero). Nominal – items are classified into categories that have no quantitative relationship to one another.

24 24 Choosing a test statistic Level of measurement of dependent variable One Independent Variable Two Independent Variables Two Treatments More Than Two Treatments Factorial Designs Two Independent Groups Two matched groups (or within subjects) Multiple independent groups Multiple matched groups (or within subjects) Independent groups Matched groups (or within sujects) Independent groups and matched groups (or between subjects and within subjects Interval or ratio t test for independent groups t test for matched groups One-way ANOVA One-way ANOVA (repeated measures) Two-way ANOVA Two-way ANOVA (repeated measures) Two-way ANOVA (mixed) ordinal Mann- Whitney U test Wilcoxon test Kruskal- Wallis test Friedman test test Nominal Chi square test TABLE 14-2 Selecting a possible statistical test by number of independent variables and level of measurement

25 25 Heading Error: Statistical Analysis t test for Independent Groups X 1 – X 2 t obs = (n 1 + n 2 – 2) ( (n 1 – 1)s 2 1 + (n 2 – 1)s 2 2 (n 1 – 1) s 2 1 + (n 2 – 1) s 2 2 )() 1 1 + n 1 n 2  22.74 – 9.08 t obs = (8 + 8 – 2) ( (8–1)236.29+(8–1)22.71 )() 1 1 + 8 8  1) Lay out Formula 2) Plug in Values

26 26 Heading Error: Statistical Analysis t test for Independent Groups t obs = 8) Divide the numerator by the denominator. X 1 – X 2 t obs = (n 1 + n 2 – 2) ( (n 1 – 1)s 2 1 + (n 2 – 1)s 2 2 (n 1 – 1) s 2 1 + (n 2 – 1) s 2 2 )() 1 1 + n 1 n 2  Formula 13.66 5.69 t obs = 2.40

27 27 Determining significance 1.Was the hypothesis directional or nondirectional? 2.What was the significance level? 3.How many degrees of freedom do we have? Degrees of freedom (df)– the number of members in a set of data that can vary or change value without changing the value of a known statistic for those data.

28 28 1.Was the hypothesis directional or nondirectional? Nondirectional, so two- tailed. 2.What was the significance level? p <.05 3.How many degrees of freedom do we have? 14 Answers to the questions Look on page 531 of Myers & Hansen to find the critical value of t…

29 29 Answers to the questions Or you could just go on-line… e.g., http://www.psychstat.missouristate.edu/introbook/tdist.htm Or you could just go on-line… e.g., http://www.psychstat.missouristate.edu/introbook/tdist.htm

30 30 Heading Error: Statistical Analysis t test for Independent Groups t obs = 8) Divide the numerator by the denominator. X 1 – X 2 t obs = (n 1 + n 2 – 2) ( (n 1 – 1)s 2 1 + (n 2 – 1)s 2 2 (n 1 – 1) s 2 1 + (n 2 – 1) s 2 2 )() 1 1 + n 1 n 2  Formula 13.66 5.69 t obs = 2.40 t crit = 2.145 p <.05, two-tailed

31 31 Compare to our computer output from SPSS t obs = 8) Divide the numerator by the denominator. X 1 – X 2 t obs = (n 1 + n 2 – 2) ( (n 1 – 1)s 2 1 + (n 2 – 1)s 2 2 (n 1 – 1) s 2 1 + (n 2 – 1) s 2 2 )() 1 1 + n 1 n 2  Formula 13.66 5.69 t obs = 2.40 t crit = 2.145 p <.05, two-tailed

32 32 Decision: Reject the null hypothesis Are we done? Conclusion - How much importance should we attach to this finding? - Was the effect just barely significant (p<.05)? - What if the sig level was, p<.0001? Would this be a larger effect?

33 33 Assess the quality of the Experiment 1)Were control procedures adequate? 2)Were variables defined appropriately? 3)Is a Type I error likely? Answers to the questions The t test is a robust statistic… Means that assumptions can be violated without changing the rate of type I or type II error.

34 34 Effect size Convert t to a correlation coefficient r =r =r =r = t2t2t2t2  t 2 +df r =r =r =r = (2.40) 2  (2.40) 2 +14 r =.54 r 2 =.15 According to Cohen (1988), r ≥.50 is considered a large effect (.30 is a moderate effect and below.30 is a small effect). The r 2 of.15 indicates that the IV accounts for 15% of the variability observed in the DV. Online site for effect size calculator: http://web.uccs.edu/lbecker/Psy590/escalc3.htm

35 35 Effect size Convert t to a correlation coefficient r =r =r =r = t2t2t2t2  t 2 +df r =r =r =r = (2.40) 2  (2.40) 2 +14 r =.54 r 2 =.15 Online site for effect size calculator: http://web.uccs.edu/lbecker/Psy590/escalc3.htm


Download ppt "1 Why do we need statistics? A.To confuse students B.To torture students C.To put the fear of the almighty in them D.To ruin their GPA, so that they don’t."

Similar presentations


Ads by Google