Chapter 20 t - Test for Two Matched Samples
Vodka & Whiskey The researcher randomly selected 10 Cranberry vodka drinkers and 7 Boilermaker drinkers. The  vodka group had a mean drunkenness of 52  and the  whiskey group had a mean drunkenness of 39 . Previous Experiment Previous Decision The drunkenness of vodka drinkers and whiskey drinkers is different Question : Who typically drinks Cranberry Vodkas and who drinks Boilermakers? 2nd Question Who typically weighs more? 3rd Question : Was the difference in the previous experiment due to different alcohol or different weights? Women Men Lighter Heavier
Two Independent Sample  t -Test Factors that affect the  standard error  ( s   1 -  2 ) Population Standard Deviation Random Sampling and Sample Size Variable estimates of the   Variability between the selected samples Possibility that the samples differ in the other factors that affect the measure. e.g. Mostly light group vs mostly heavy group Always a problem with random selection
Two Independent Sample  t -Test Factors that affect the  standard error  ( s   1 -  2 ) Population Standard Deviation Random Sampling and Sample Size Variable estimates of the   Variability between the selected samples Always going to affect the error Can we minimize this factor by forcing the selected samples to be similar?
Matched Samples  t -Test Reducing a Two-Sample test into a Single Sample Comparison  (Single Sample  t -Test)
Matched Samples  t -Test FAQ What does matching mean? An observation in one sample is paired with an observation in the other sample?
Matched Samples  t -Test FAQ How does matching happen? Pick a matching variable Must be correlated to the  dependent measure e.g., weight and drunkenness Randomly select 2 participants who are equal on the matching factor Randomly assign 1 participant to each group Vodka: 110lb, 150lb, 190lb Whiskey: 110lb, 150lb, 190lb
Matched Samples  t -Test FAQ What  should  matching do? Reduce the between sample variability e.g., Affected by weight the same Reduce the  standard error . What is special about the matched samples  t -test? Mathematically reduces the  standard error . Brings the  critical scores  closer to zero Makes the test more sensitive.
Matched Samples  t -Test FAQ What does the test assume? Populations have a normal shape …. or at least the sampling distributions are normal Homogeneity of Variance Population have the same variability Matching variable is correlated to the  dependent measure
Matched Samples  t -Test FAQ What is a difference score? Actual dependent measure used by the test. It is the difference between the paired scores D 110lb  = Vodka 110lb  - Whiskey 110lb What is the null hypothesis? The groups are equal Vodka 110lb  - Whiskey 110lb  = 0  D  = 0
Matched Samples  t -Test FAQ Any other requirements? Equal sample sizes!!!!!
Hypothesis Test 11 Vodka & Whiskey  Matched Samples t-Test
Vodka/Whiskey: Matched  t -Test Are all types of alcohol the same, even if the proofs are the same? This question was raised by a researcher who had observed vodka drinkers and noticed that they seemed to get drunk faster than whiskey drinkers.  To test whether whiskey is the same or different than vodka, the researcher decided to compare people who drank 3 Cranberry Vodka’s to people to drank 3 Boilermakers. Since weight is known to affect drunkenness, the researcher has matched the samples to make sure that weight is evenly distributed between the group (on next slide).What will the researcher conclude at a .01 level of significance.
Vodka/Whiskey: Matched  t -Test Step 0) Convert to Difference Scores 19 25 210 20 24 190 31 30 170 36 34 150 42 47 130 63 52 110 Whiskey Vodka Weight +6 +4 -1 -2 + 5 - 11 D
Vodka/Whiskey: Matched  t -Test Step 0) Convert to Difference Scores =   D  /   n dif =    /   6 =  .16 +6 +4 -1 -2 + 5 - 11 D D   (D 2  s D   =  D) 2 n dif n dif   - 1
Vodka/Whiskey: Matched  t -Test Step 0) Convert to Difference Scores =   D  /   n dif =    /   6 =  .16 =  6.37 +6 +4 -1 -2 + 5 - 11 D D   (D 2  s D   =  D) 2 n dif n dif   - 1   s D   =  ) 2 6 6   - 1 36 16 1 4 25 121 D 2
Vodka/Whiskey: Matched  t -Test Step 1) Rewrite the research question “ Does the mean drunkenness of the vodka population equal the mean drunkenness of the whiskey population.” Step 2) Write the statistical hypotheses H 0 :   D = 0  H 1 :   D    0
Vodka/Whiskey: Matched  t -Test Is the mean of the vodka pop. the same as the whiskey pop? Hypothesis H 0 :   D =0  H 1 :   D    0 Step 3) Form Decision Rule Draw Normal Curve Shade in   Mark  Rejection Region(s) Determine  Critical Scores Write conditions for rejection H 0 A)   =.16 s D =6.37 n dif  = 6  df dif  = 5     = .01 C) Reject H 0   Fail to reject H 0   Reject H 0 B) .005 .005 t( 5 ) crit =  -4.032 t( 5 ) crit = +4.032 Decision Rule:  Reject H 0 t obt  < -4.032 t obt  > +4.032 E) D
Vodka/Whiskey: Matched  t -Test Is the mean of the vodka pop. the same as the whiskey pop? Hypothesis H 0 :   D =0  H 1 :   D    0  D  =   hyp =  0 =  2.6   =.16 s D =6.37 n dif  = 6  df dif  = 5     = .01 Decision Rule:  Reject H 0 t obt  < -4.032 t obt  > +4.032 E) D  = 6   = 2.45 s D s D = n dif 
Vodka/Whiskey: Matched  t -Test Is the mean of the vodka pop. the same as the whiskey pop? Hypothesis H 0 :   D =0  H 1 :   D    0 Based upon a sampling distribution with    &  =  2.6 Step 4) Calculate Test Statistic  D  =   = .06   =.16 s D =6.37 n dif  = 6  df dif  = 5     = .01 Decision Rule:  Reject H 0 t obt  < -4.032 t obt  > +4.032 E) D s D t obt  =  D s D D - t obt  = .16 - 0 2.6
Vodka/Whiskey: Matched  t -Test Step 6) Interpret Decision Step 5) Make Decision We have no evidence to suggest that the drunkenness of vodka drinkers is different from whiskey drinkers. Is the mean of the vodka pop. the same as the whiskey pop? Hypothesis H 0 :   D =0  H 1 :   D    0 Based upon a sampling distribution with    &  =  2.6  D  =   t obt  = .06 Fail to Reject H 0   =.16 s D =6.37 n dif  = 6  df dif  = 5     = .01 Decision Rule:  Reject H 0 t obt  < -4.032 t obt  > +4.032 E) D s D
Why Matching Works! Why can we assume the standard error is reduced ( if the matching variable is correlated with the dependent measure )
Why Matching Works
When Matching Doesn’t Work
Matched Sample  t -Test Matched Sample  t -Test and the correlation assumption. The test does not know if the matching variable is correlated Assumes it is correlated because you selected it Drops the standard error estimate If the matching variable is not correlated… Standard error has not actually decreased ….. but the test lowered it anyway.
Repeated Measures The Ultimate Matching
Repeated Measures Repeated Measures Using the same participants in both conditions Concern : Carry-over Effects Spill-Over Effects of drugs Linger Practice effects Second time the participant has done the task Counter-balancing Some participants given the conditions in reverse order
To match, or not to match… That is the question……
Matched vs Independent Tests Which test is more sensitive? Smaller Larger Standard Error Less More Degrees of Freedom Matched Independent
Matched vs Independent Tests Which test is more sensitive? Brings in Critical Values Spreads out Critical Values Standard Error Spreads out Critical Values Brings in Critical Values Degrees of Freedom Matched Independent …  has a much greater effect on critical scores?
Matched vs Independent Tests Which test is more sensitive? Brings in a lot Spreads out a lot Standard Error Spreads out a little Brings in a little Degrees of Freedom Matched Independent …  has a much greater effect on critical scores? More Sensitive
Two Sample Tests Final Notes
Two Sample Tests One-Tailed Tests Make the population with the larger hypothesized mean population 1 Always an upper-critical test Confidence Intervals Confident that the difference between the population means is within the interval …. not about the value of the population means

香港六合彩

  • 1.
    Chapter 20 t- Test for Two Matched Samples
  • 2.
    Vodka & WhiskeyThe researcher randomly selected 10 Cranberry vodka drinkers and 7 Boilermaker drinkers. The vodka group had a mean drunkenness of 52 and the whiskey group had a mean drunkenness of 39 . Previous Experiment Previous Decision The drunkenness of vodka drinkers and whiskey drinkers is different Question : Who typically drinks Cranberry Vodkas and who drinks Boilermakers? 2nd Question Who typically weighs more? 3rd Question : Was the difference in the previous experiment due to different alcohol or different weights? Women Men Lighter Heavier
  • 3.
    Two Independent Sample t -Test Factors that affect the standard error ( s  1 -  2 ) Population Standard Deviation Random Sampling and Sample Size Variable estimates of the  Variability between the selected samples Possibility that the samples differ in the other factors that affect the measure. e.g. Mostly light group vs mostly heavy group Always a problem with random selection
  • 4.
    Two Independent Sample t -Test Factors that affect the standard error ( s  1 -  2 ) Population Standard Deviation Random Sampling and Sample Size Variable estimates of the  Variability between the selected samples Always going to affect the error Can we minimize this factor by forcing the selected samples to be similar?
  • 5.
    Matched Samples t -Test Reducing a Two-Sample test into a Single Sample Comparison (Single Sample t -Test)
  • 6.
    Matched Samples t -Test FAQ What does matching mean? An observation in one sample is paired with an observation in the other sample?
  • 7.
    Matched Samples t -Test FAQ How does matching happen? Pick a matching variable Must be correlated to the dependent measure e.g., weight and drunkenness Randomly select 2 participants who are equal on the matching factor Randomly assign 1 participant to each group Vodka: 110lb, 150lb, 190lb Whiskey: 110lb, 150lb, 190lb
  • 8.
    Matched Samples t -Test FAQ What should matching do? Reduce the between sample variability e.g., Affected by weight the same Reduce the standard error . What is special about the matched samples t -test? Mathematically reduces the standard error . Brings the critical scores closer to zero Makes the test more sensitive.
  • 9.
    Matched Samples t -Test FAQ What does the test assume? Populations have a normal shape …. or at least the sampling distributions are normal Homogeneity of Variance Population have the same variability Matching variable is correlated to the dependent measure
  • 10.
    Matched Samples t -Test FAQ What is a difference score? Actual dependent measure used by the test. It is the difference between the paired scores D 110lb = Vodka 110lb - Whiskey 110lb What is the null hypothesis? The groups are equal Vodka 110lb - Whiskey 110lb = 0  D = 0
  • 11.
    Matched Samples t -Test FAQ Any other requirements? Equal sample sizes!!!!!
  • 12.
    Hypothesis Test 11Vodka & Whiskey  Matched Samples t-Test
  • 13.
    Vodka/Whiskey: Matched t -Test Are all types of alcohol the same, even if the proofs are the same? This question was raised by a researcher who had observed vodka drinkers and noticed that they seemed to get drunk faster than whiskey drinkers. To test whether whiskey is the same or different than vodka, the researcher decided to compare people who drank 3 Cranberry Vodka’s to people to drank 3 Boilermakers. Since weight is known to affect drunkenness, the researcher has matched the samples to make sure that weight is evenly distributed between the group (on next slide).What will the researcher conclude at a .01 level of significance.
  • 14.
    Vodka/Whiskey: Matched t -Test Step 0) Convert to Difference Scores 19 25 210 20 24 190 31 30 170 36 34 150 42 47 130 63 52 110 Whiskey Vodka Weight +6 +4 -1 -2 + 5 - 11 D
  • 15.
    Vodka/Whiskey: Matched t -Test Step 0) Convert to Difference Scores =  D / n dif =  / 6 = .16 +6 +4 -1 -2 + 5 - 11 D D   (D 2  s D =  D) 2 n dif n dif - 1
  • 16.
    Vodka/Whiskey: Matched t -Test Step 0) Convert to Difference Scores =  D / n dif =  / 6 = .16 = 6.37 +6 +4 -1 -2 + 5 - 11 D D   (D 2  s D =  D) 2 n dif n dif - 1   s D =  ) 2 6 6 - 1 36 16 1 4 25 121 D 2
  • 17.
    Vodka/Whiskey: Matched t -Test Step 1) Rewrite the research question “ Does the mean drunkenness of the vodka population equal the mean drunkenness of the whiskey population.” Step 2) Write the statistical hypotheses H 0 :  D = 0 H 1 :  D  0
  • 18.
    Vodka/Whiskey: Matched t -Test Is the mean of the vodka pop. the same as the whiskey pop? Hypothesis H 0 :  D =0 H 1 :  D  0 Step 3) Form Decision Rule Draw Normal Curve Shade in  Mark Rejection Region(s) Determine Critical Scores Write conditions for rejection H 0 A)   =.16 s D =6.37 n dif = 6  df dif = 5   = .01 C) Reject H 0 Fail to reject H 0 Reject H 0 B) .005 .005 t( 5 ) crit = -4.032 t( 5 ) crit = +4.032 Decision Rule: Reject H 0 t obt < -4.032 t obt > +4.032 E) D
  • 19.
    Vodka/Whiskey: Matched t -Test Is the mean of the vodka pop. the same as the whiskey pop? Hypothesis H 0 :  D =0 H 1 :  D  0  D =  hyp = 0 = 2.6   =.16 s D =6.37 n dif = 6  df dif = 5   = .01 Decision Rule: Reject H 0 t obt < -4.032 t obt > +4.032 E) D  = 6   = 2.45 s D s D = n dif 
  • 20.
    Vodka/Whiskey: Matched t -Test Is the mean of the vodka pop. the same as the whiskey pop? Hypothesis H 0 :  D =0 H 1 :  D  0 Based upon a sampling distribution with  & = 2.6 Step 4) Calculate Test Statistic  D =  = .06   =.16 s D =6.37 n dif = 6  df dif = 5   = .01 Decision Rule: Reject H 0 t obt < -4.032 t obt > +4.032 E) D s D t obt =  D s D D - t obt = .16 - 0 2.6
  • 21.
    Vodka/Whiskey: Matched t -Test Step 6) Interpret Decision Step 5) Make Decision We have no evidence to suggest that the drunkenness of vodka drinkers is different from whiskey drinkers. Is the mean of the vodka pop. the same as the whiskey pop? Hypothesis H 0 :  D =0 H 1 :  D  0 Based upon a sampling distribution with  & = 2.6  D =  t obt = .06 Fail to Reject H 0   =.16 s D =6.37 n dif = 6  df dif = 5   = .01 Decision Rule: Reject H 0 t obt < -4.032 t obt > +4.032 E) D s D
  • 22.
    Why Matching Works!Why can we assume the standard error is reduced ( if the matching variable is correlated with the dependent measure )
  • 23.
  • 24.
  • 25.
    Matched Sample t -Test Matched Sample t -Test and the correlation assumption. The test does not know if the matching variable is correlated Assumes it is correlated because you selected it Drops the standard error estimate If the matching variable is not correlated… Standard error has not actually decreased ….. but the test lowered it anyway.
  • 26.
    Repeated Measures TheUltimate Matching
  • 27.
    Repeated Measures RepeatedMeasures Using the same participants in both conditions Concern : Carry-over Effects Spill-Over Effects of drugs Linger Practice effects Second time the participant has done the task Counter-balancing Some participants given the conditions in reverse order
  • 28.
    To match, ornot to match… That is the question……
  • 29.
    Matched vs IndependentTests Which test is more sensitive? Smaller Larger Standard Error Less More Degrees of Freedom Matched Independent
  • 30.
    Matched vs IndependentTests Which test is more sensitive? Brings in Critical Values Spreads out Critical Values Standard Error Spreads out Critical Values Brings in Critical Values Degrees of Freedom Matched Independent … has a much greater effect on critical scores?
  • 31.
    Matched vs IndependentTests Which test is more sensitive? Brings in a lot Spreads out a lot Standard Error Spreads out a little Brings in a little Degrees of Freedom Matched Independent … has a much greater effect on critical scores? More Sensitive
  • 32.
    Two Sample TestsFinal Notes
  • 33.
    Two Sample TestsOne-Tailed Tests Make the population with the larger hypothesized mean population 1 Always an upper-critical test Confidence Intervals Confident that the difference between the population means is within the interval …. not about the value of the population means