Independent-Samples t Test Lecture PDF

Independent-Samples t test Used to test for a difference between two groups when using a between-subjects design with independent samples  Single-Sample z-Test t-Test  Randomly selected  Randomly selected sampl...

Independent-Samples t test Used to test for a difference between two groups when using a between-subjects design with independent samples  Single-Sample z-Test t-Test  Randomly selected  Randomly selected sample sample  DV normally distributed  DV normally distributed  DV measured using  DV measured using ratio or interval scale ratio or interval scale  mean of the population  mean of the population is known is known  SD of the population  SD of the population is known is not known and must be estimated  Single-Sample Independent-Samples t-Test t-Test  Randomly selected  Randomly selected sample sample  DV normally distributed  DV normally distributed  DV measured using  DV measured using ratio or interval scale ratio or interval scale  mean of the population  Homogeneity of is known variance  SD of the population is not known and must be estimated  General Model for z-Test and Single-Sample t-Test Original H0 Population Sample HA Treated Population General Model for Independent-Samples t-Test Sample A Population A H0 Sample B Population B H 0 : 1 - 2 = 0 General Model for Independent-Samples t-Test Sample A Population A HA Sample B Population B H A : 1 - 2  0  Sampling Distribution of the Difference Between the Means S X1  X 2 f X1 X 2 X1 X 2 μ X1 X 2 X1 X 2  Sampling Distribution of Difference Between the Means To create this sampling distribution:  Select 2 random samples from one population Each sample is the same size as the N of our groups  Compute the sample mean for each sample  Subtract one sample mean from the other and plot the difference  Do this an infinite # of times  Standard Error of the Difference between the Means S X1  X 2 The average distance between the mean of the sampling distribution (of the difference between the means) and all of the differences between the means plotted in the sampling distribution of the differences between the means.  How much difference should you expect between the sample means even if your treatment has no effect?  t Tests Formulas Single-Sample t-Test sample mean  population mean tobt  estimated standard error (of the mean) Independent-Samples t-Test sample diff. btwn the means  population diff. btwn. the means tobt  estimated standard error (of the difference between th e means)  Formula Definitional Formulas Single-Sample Independent-Samples t-Test t-Test X  ( X 1  X 2 )  ( 1   2 ) tobt  tobt  sX s X1  X 2  Single-Sample t-Test Step 1: s 2   (X  X) 2 X Step 2: n 1 2 Estimated variance of the s population (definitional formula) sX  X Step 3: n X  Estimated standard error of the mean tobt  sX  Single-Sample Independent- Samples t-Test t-Test s 2   (X  X ) 2 s 2   (X  X ) 2 X X n 1 n 1 Step 1:  calculate the estimated variance of the population  calculate the estimated variance of the population for each group Pooled Variance Step 1a:  Calculate the pooled variance 2 SS1  SS 2 s pool  df1  df 2  Pooled Variance  Equal Sample Sizes SS1  SS 2 SS1 = 50 SS2 = 30 2 s pool  n1 = 6 n2 = 6 df1  df 2 2 50  30 80 s pool   8 55 10 SS 1 SS 2 50 30 [  ] / 2 [  ] / 2  df 1 df 2 5 5 Average of s12 [10  6] / 2 16 / 2 8 and s22 Pooled Variance  Unequal Sample Sizes SS1  SS 2 SS1 = 20 SS2 = 48 2 s pool  n1 = 3 n2 = 9 df1  df 2 2  20  48 68  6.8 s pool 2 8 10 SS 1 SS 2 20 48 [  ] / 2 [  ] / 2  Average of s12 df 1 df 2 2 8 and s22 [10  6] / 2 16 / 2 8  Single-Sample Independent- Samples t Test t Test 2 s 2 2 s s     pool pool sX  X s X1  X 2  n   n  1 n2  Step 2:  calculate the estimated standard error of the mean  calculate the estimated standard error of the difference between the means (standard error of the difference)  Single-Sample Independent- Samples t-Test t-Test 2 s 2 1 1 sX  X s X1  X 2  s pool    n  n1 n2  Step 2:  calculate the estimated standard error of the mean  calculate the estimated standard error of the difference between the means (standard error of the difference)  Single-Sample Independent- Samples t-Test t-Test X  ( X 1  X 2 )  ( 1   2 ) tobt  tobt  sX s X1  X 2 Step 3:  calculate tobt  calculate tobt  Single-Sample Independent- Samples t-Test t-Test X  ( X 1  X 2 )  ( 1   2 ) tobt  tobt  sX s X1  X 2 Step 3:  calculate tobt  calculate tobt Hypothesis Testing with Two Independent Samples Step 1. State the hypotheses (two- tailed) A. Is it a one-tailed or two-tailed test? Two-tailed B. Research hypotheses  Alternative hypothesis: There is a difference between the control group and the experimental group.  Null hypothesis: There is no difference between the control group and the experimental group. C. Statistical hypotheses: HA: 1 - 2  0 which is equivalent to 1  2 H0: 1 - 2 = 0 which is equivalent to 1 = 2 The HA and H0 Hypotheses  The HA says that there is a difference between the groups, so your difference is NOT zero  The H0 says that there is NOT a difference, so your difference equals zero  You can put the control group or the experimental group as group 1 in your equations, but you HAVE TO BE CONSISTENT  You should substitute abbreviated names based on the conditions instead of 1 and 2 as subscripts Step 1. State the hypotheses. A. Is it a one-tailed or two-tailed test? One-tailed Step 1. State the hypotheses (one- tailed) B. Research hypotheses  Alternative hypothesis: The experimental group will perform better than the control group. The experimental group’s scores will be lower than the control group’s score.  Null hypothesis: The experimental group will perform the same as or worse than the control group. The experimental group’s scores will be the same as or higher than the control group’s scores. C. Statistical hypotheses: HA: experimental - control > 0 experimental - control < 0 H0: experimental - control < 0 experimental - control > 0 Step 1. State the hypotheses. A. Is it a one-tailed or two-tailed test? One-tailed B. Research hypotheses  Alternative hypothesis: Participants who eat peppermint will score higher than those who don’t eat peppermint on the digit recall test.  Null hypothesis: Participants who eat peppermint will score the same as or lower than those who don’t eat peppermint on the digit recall test. C. Statistical hypotheses: HA: peppermint - no peppermint > 0 H0: peppermint - no peppermint < 0 Step 2. Set the significance level   =.05. Determine tcrit. Factors that must be known to find tcrit 1. Is it a one-tailed or a two-tailed test?  one-tailed 2. What is the alpha level? .05 3. What are the degrees of freedom?  df = ? Degrees of Freedom Single-Sample Independent-Samples t-Test t-Test df = (n – 1) df = (n1 – 1) + (n2 – 1) = n1 + n2 – 2  Step 3. Select and compute the appropriate statistical test. Step 1: Step 1a: s 2   (X  X ) 2 s 2 pool  SS1  SS 2 X n 1 df1  df 2 Step 2: Step 3:  s 2pool s 2pool  ( X 1  X 2 )  ( 1   2 ) s X1  X 2     tobt   n  1 n 2   s X1  X 2 Step 4. Make a decision.  Determine whether the value of the test statistic is in the critical region. Draw a picture. tcrit = ??? Step 4. Make a decision.  If +tobt > +tcrit OR if -tobt < -tcrit  Reject Ho  If -tcrit < tobt < +tcrit  Retain Ho -tcrit +tcrit REJECT H0 RETAIN H0 REJECT H0 Step 5. Report the statistical results.  Reject H0: t(df) = tobt, p <.05  Retain H0: t(df) = tobt, p >.05 Step 6: Write a conclusion.  State the relationship between the IV and the DV in words, ending with the statistical results.  General format: Members of the first group (M = xx.xx) did/did not score lower/higher/differently than members of the second group (M = xx.xx), t(df) = tobt, p < >.05. Hypothesis Testing with Two Independent Samples An Example An Example  Research Question: Are students who calculate statistics by hand better able to select the appropriate statistical test to use than students who do not calculate statistics by hand (who use SPSS)?  Assume that past research has consistently shown that students who calculate statistics by hand are better, so we decide to generate a directional (one-tailed) hypothesis. Step 1. State the hypotheses. A. Is it a one-tailed or two-tailed test?  One-tailed B. Research hypotheses  Alternative hypothesis: Students who calculate statistics by hand are better able to select the appropriate statistical test to use than students who do not calculate statistics by hand.  Null hypothesis: Students who calculate statistics by hand are not better (i.e., are no different from or are less able) to select the appropriate statistical test than students who do not calculate stats by hand. C. Statistical hypotheses:  HA: hand - SPSS > 0  H0: hand - SPSS < 0 Step 2. Set the significance level   =.05. Determine tcrit. 1. Is it a one-tailed or a two-tailed test?  one-tailed 2. What is the alpha level? .05 3. What are the degrees of freedom? n1 = 5; n2 = 5  df = (n – 1) + (n – 1) 1 2 = (5 -1) + (5-1) =4+4 =8 tcrit = 1.860 Step 3. Select and compute the appropriate statistic. Independent-Samples t-Test  Hand- Calculation Scores (X1) X X–X (X – X )2 8 7 10 9 8 ∑X1= n= ∑(X –X )2 X1  =  Hand- Calculation Scores (X1) X X–X (X – X )2 8 8.4 -.4.16 7 8.4 -1.4 1.96 10 8.4 1.6 2.56 9 8.4.6.36 8 8.4 -.4.16 ∑X1= 42 n=5 ∑(X –X )2 X 1 8.4 = 5.2  SPSS Scores (X2) X X–X (X – X )2 6 5 7 8 6 ∑X2= n= ∑(X –X )2 X2  =  SPSS Scores (X2) X X–X (X – X )2 6 6.4 -.4.16 5 6.4 -1.4 1.96 7 6.4.6.36 8 6.4 1.6 2.56 6 6.4 -.4.16 ∑X2= 32 n=5 ∑(X –X )2 X 2 6.4 = 5.2 1) calculate the estimated variance of each population s 2   (X  X ) 2 X n 1 1a) calculate the pooled variance 2 SS1  SS 2 s pool  df1  df 2  2 SS1  SS 2 s pool  df1  df 2 5.2  5.2  44 10.4  8 2 s pool 1.3 2) calculate the estimated standard error of the difference between the means 2 2 s s     pool pool s X1  X 2   n n   1 2    s 2pool s 2pool  s X1  X 2      n n2   1   1.3 1.3       5 5   2.6      5  .52 s X 1  X 2 .721 3) calculate tobt ( X 1  X 2 )  ( 1   2 ) tobt  s X1  X 2  ( X 1  X 2 )  ( 1   2 ) tobt  s X1  X 2 (8.4  6.4)  (0) .721 2 .721 tobt 2.77 Step 4. Make a decision.  If +tobt > +tcrit  Reject Ho  If tobt < +tcrit  Retain Ho tcrit = + 1.860 tobt = 2.77 Step 5. Report the statistical results. t(8) = 2.77, p <.05 Step 6: Write a conclusion.  State the relationship between the IV and the DV in words: Students who calculate statistics by hand (M = 8.4) are significantly better at selecting the appropriate statistical test to use than students who do not calculate statistics by hand (M = 6.4), t(8) = 2.77, p <.05. Step 7. Compute the estimated d. X1  X 2 estimated d  2 s pool Step 7. Compute the estimated d. X1  X 2 estimated d  2 s pool 8.4  6.4 2   1.75 1.3 1.140 Percentage of Variance Explained (r2)  Describingthe Strength of the Relationship 2 2 t r  2 t  df  Step 8. Compute r2 and write a conclusion. 2 2 t r  2 t  df 2 (2.77) 7.673 7.673  2   .4896 (2.77)  8 7.673  8 15.673 Step 8. Compute r2 and write a conclusion.  Approximately 49% of the variance in students’ ability to select the appropriate statistical test can be accounted for by their group membership (i.e., the way they learned statistics).  The way students learned statistics can account for 48.96% of the variance in their ability to select the appropriate statistical test.