Chapter 7 Inference Concerning Populations (Numeric Responses)

Inference for Population Mean Practical Problem: Sample mean has sampling distribution that is Normal with mean  and standard deviation  /  n (when the data are normal, and approximately so for large samples).  is unknown. Have an estimate of , s obtained from sample data. Estimated standard error of the sample mean is: When the sample is SRS from N(  then the t-statistic (same as z- with estimated standard deviation) is distributed t with n-1 degrees of freedom

Family of t-distributions Symmetric, Mound-shaped, centered at 0 (like the standard normal (z) distribution Indexed by degrees of freedom  )  the number of independent observations (deviations) comprising the estimated standard deviation. For one sample problems = n-1 Have heavier tails (more probability over extreme ranges) than the z-distribution Converge to the z-distribution as gets large Tables of critical values for certain upper tail probabilities are available (inside back cover of text)

Probability DegreesofFreedomDegreesofFreedom CriticalValuesCriticalValues Critical Values

One-Sample Confidence Interval for  SRS from a population with mean  is obtained. Sample mean, sample standard deviation are obtained Degrees of freedom are = n-1, and confidence level C are selected Level C confidence interval of form: Procedure is theoretically derived based on normally distributed data, but has been found to work well regardless for large n

1-Sample t-test (2-tailed alternative) 2-sided Test: H 0 :  =  0 H a :    0 Decision Rule (t * obtained such that P(t(n-1)  t * )=  /2) : –Conclude  >  0 if Test Statistic (t obs ) is greater than t * –Conclude  <  0 if Test Statistic (t obs ) is less than -t * –Do not conclude Conclude    0 otherwise P-value: 2P(t(n-1)  |t obs |) Test Statistic:

P-value (2-tailed test)

1-Sample t-test (1-tailed (upper) alternative) 1-sided Test: H 0 :  =  0 H a :  >  0 Decision Rule (t * obtained such that P(t(n-1)  t * )=  ) : –Conclude  >  0 if Test Statistic (t obs ) is greater than t * –Do not conclude  >  0 otherwise P-value: P(t(n-1)  t obs ) Test Statistic:

P-value (Upper Tail Test)

1-Sample t-test (1-tailed (lower) alternative) 1-sided Test: H 0 :  =  0 H a :  <  0 Decision Rule (t * obtained such that P(t(n-1)  t * )=  ) : –Conclude  <  0 if Test Statistic (t obs ) is less than -t * –Do not conclude  <  0 otherwise P-value: P(t(n-1)  t obs ) Test Statistic:

P-value (Lower Tail Test)

Example: Mean Flight Time ATL/Honolulu Scheduled flight time: 580 minutes Sample: n=31 flights 10/2004 (treating as SRS from all possible flights Test whether population mean flight time differs from scheduled time H 0 :  = 580 H a :   580 Critical value (2-sided test,  = 0.05, n-1=30 df): t * =2.042 Sample data, Test Statistic, P-value:

Paired t-test for Matched Pairs Goal: Compare 2 Conditions on matched individuals (based on similarities) or the same individual under both conditions (e.g. before/after studies) Obtain the difference for each pair/individual Obtain the mean and standard deviation of the differences Test whether the true population means differ (e.g. H 0 :  D = 0) Test treats the differences as if they were the raw data

Test Concerning  D Null Hypothesis : H 0 :  D =  0 (almost always 0) Alternative Hypotheses : –1-Sided: H A :  D >  0 –2-Sided : H A :  D   0 Test Statistic:

Test Concerning  D Decision Rule: (Based on t-distribution with =n-1 df) 1-sided alternative If t obs  t * ==> Conclude  D    If t obs Do not reject  D    2-sided alternative If t obs  t * ==> Conclude  D    If t obs  -t * ==> Conclude  D    If -t * Do not reject  D    Confidence Interval for  D

Example Antiperspirant Formulations Subjects - 20 Volunteers’ armpits Treatments - Dry Powder vs Powder-in-Oil Measurements - Average Rating by Judges –Higher scores imply more disagreeable odor Summary Statistics (Raw Data on next slide): Source: E. Jungermann (1974)

Example Antiperspirant Formulations

Evidence that scores are higher (more unpleasant) for the dry powder (formulation 1)

Comparing 2 Means - Independent Samples Goal: Compare responses between 2 groups (populations, treatments, conditions) Observed individuals from the 2 groups are samples from distinct populations (identified by (  1,  1 ) and (  2,  2 )) Measurements across groups are independent (different individuals in the 2 groups Summary statistics obtained from the 2 groups:

Sampling Distribution of Underlying distributions normal  sampling distribution is normal Underlying distributions nonnormal, but large sample sizes  sampling distribution approximately normal Mean, variance, standard deviation:

t-test when Variances are estimated Case 1: Population Variances not assumed to be equal (  1 2  2 2 ) Approximate degrees of freedom –Calculated from a function of sample variances and sample sizes (see formula below) - Satterthwaite’s approximation –Smaller of n 1 -1 and n 2 -1 Estimated standard error and test statistic for testing H 0 :  1 =  2 :

t-test when Variances are estimated Case 2: Population Variances assumed to be equal (  1 2 =  2 2 ) Degrees of freedom: n 1 +n 2 -2 Estimated standard error and test statistic for testing H 0 :  1 =  2 :

Example - Maze Learning (Adults/Children) Groups: Adults (n 1 =14) / Children (n 2 =10) Outcome: Average # of Errors in Maze Learning Task Raw Data on next slide Conduct a 2-sided test of whether mean scores differ Construct a 95% Confidence Interval for true difference Source: Gould and Perrin (1916)

Example - Maze Learning (Adults/Children)

Example - Maze Learning Case 1 - Unequal Variances H 0 :      H A :      0 (  = 0.05) No significant difference between 2 age groups Note: Alternative would be to use 9 df (10-1)

Example - Maze Learning Case 2 - Equal Variances H 0 :      H A :      0 (  = 0.05) No significant difference between 2 age groups

SPSS Output

C% Confidence Interval for  1 -  2