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Introduction to the t statistic • Overview – The previous chapter used a z statistic to determine whether an observed mean differed from the population mean – Z = (M - )/M – Limitation: necessary to know the population SD – a t-statistic can be used in situations in which the population SD is not known Introduction to the t statistic • The t statistic – The t statistic is similar to the z statistic; however, unlike the z statistic, it estimates from the data the population variability – Recall SS = ∑ (X – sample mean)2 – Sample variance = s2 = SS/ (n-1) – Sample standard deviation = s = √(SS/n-1) Or s = √SS/df Introduction to the t statistic • The t statistic – When the population variance is known. – Standard error = σM = σ/n.5 = √ σ2/n – Estimated standard error sM = s/n.5 = √ s2/n – Note: sM is an estimate of the standard error – Used when σ is unknown – Computed from the sample variance – Estimates distance between M (sample mean) and Introduction to the t statistic • The t statistic – The t statistic = t = (M-)/ sM – where sM = √ s2/n • Summary – the t statistic used to test hypotheses about an unknown when is unknown. Structure of t similar to z Introduction to the t statistic • Similarity of z and t statistic – Z = (M - )/M = (M - )/√ 2 /n – t statistic = t = (M-)/ sM = (M-)/ √ s2/n – Note: 2 statistics are similar except that s2 is used with t and 2 is used with z; thus, the t statistic approximates the z statistic to the extent that the sample variance approximates the population variance Introduction to the t statistic • Degrees of freedom – Degrees of freedom, df = n-1 – The larger the df the better the sample variance approximates the population variance • t distributions – By taking all possible samples of a given size (n) from a population it is possible to construct a t distribution Introduction to the t statistic • t distributions – t distributions are symmetrical just like normal distributions – As the sample size increases the shape of the t distribution becomes more similar to normal distribution (see Figure 9.1) – As this figure shows, the t distribution is flatter and more spread out – Mean of t distribution is 0 Introduction to the t statistic • t distributions – In order to test hypotheses it is necessary to determine proportions for different t values, just as one does with z distributions – Appendix B.2 has t distribution table – Note: proportion (1 or 2 tailed) is at top; left-most row has df; the body of the table has the t- distribution value Introduction to the t statistic • t distributions – E.g., study involving 20 people evaluates hypothesis with population mean memory performance is affected by a memory training program – Results find a t-value of 2.86. What proportion of scores in t-distribution have scores greater than 2.86 – Sol’n: df = 20-1 =19. look up value in table. Answer .005 Introduction to the t statistic • Hypothesis testing with t distributions – Logic is the same as with a z score. There is a known population mean. Purpose of study is to determine whether a treatment changes population mean significantly. – Step 1. state null and alternative hypothesis Population mean is unchanged as a result of treatment H0 set alpha level – Step 2. Locate critical region in t-distribution – Step 3. Collect data and compute test statistic – Step 4. Evaluate null hypothesis Introduction to the t statistic • Example – Purpose of study is to determine whether birds avoid a chamber in which eye-spot patterns have been painted. Bird placed in a chamber for 60 minutes. 1 compartment has eye-spot pattern and other (plain) compartment does not. – DV = time spent in plain compartment H0 = 30 minutes H1 ≠ 30 minutes – Alpha = .05 two tailed. why two tailed? Introduction to the t statistic • Example – Results: n=16 birds; M = 39 minutes on plain side; SS = 540 – Locate critical region Df = 15; alpha = .05, two tailed; t (15) ± 2.131 – Calculate t – t = (M-)/ sM = (M-)/ √ s2/n S2 = SS/(n-1) = 540/15 = 36 √ s2/n = √36/16 =√2.25 = 1.50 Introduction to the t statistic • Example – t = (M-)/ sM = (M-)/ √ s2/n = (39-30)/1.50 = 6.0 – Make decision Reject null hypothesis. Birds spent significantly more time in the plain compartment than in the eye-spotted compartment Introduction to the t statistic • APA format – Report descriptive statistics used M = 39 minutes, SD = 6 – Report inferential statistics – t(15) = 6.0, p < .05 – t (df) = t score, significance level Introduction to the t statistic • Importance of looking at your data – Always look at your data descriptively before looking at it inferentially – If you graph your data and the raw scores are bunched far away from the population mean you can be confident that the sample mean differs significantly from the population mean – See Figure 9.6 Introduction to the t statistic • Assumptions of the t test – 1. sample must consist of independent observations – How to ensure: use random sampling – 2. population sampled must be normal – This assumption is not critical. That is, data show that the t statistic is robust when this assumption is violated Hypothesis test with 2 independent samples • Overview – Previously we have shown that a z statistic could be used to determine whether an observed mean differed from the population mean – Z = (M - )/M – Limitation: necessary to know the population SD Hypothesis test with 2 independent samples • Overview – then we showed that a t-statistic can be used in situations in which the population SD is not known – Limitation: the particular type of t statistic can only be used when there is a single sample – In psychology and other disciplines one often wants to determine whether performance between two samples (or groups) differs Hypothesis test with 2 independent samples • Overview – This chapter will present a test that can be used in this situation – e.g., compare performance of two different training procedures administered to two different groups – e.g., compare performance of two groups trained in the same way but tested in two different ways Hypothesis test with 2 independent samples • Note: – Procedure in this chapter applies to a situation in which the data come from two separate samples – Hence this procedure is called an independent- measures research design or a between-subjects design – Subsequently we will discuss a procedure in which the comparison is made between two sets of data collected from the same sample Hypothesis test with 2 independent samples • Independent measures or between-subjects design – Experimental design that uses a different sample for each condition or treatment Hypothesis test with 2 independent samples • Goal – To determine whether there is a significant difference between the means of two different populations (or two different treatment conditions) – Notation: a subscript 1 will designate all parameters of population/treatment 1, while a subscript 2 will designate parameters of population/treatment 2 Hypothesis test with 2 independent samples • Hypothesis – H0 : 1 - 2 = 0 in other words the population means are equal – H1 : 1 - 2 ≠ 0 • Further notation – t test used to test the mean of a single sample will be referred to as the single sample t statistic – t test used to test the mean difference between two independent samples will be called the independent measures t statistic Hypothesis test with 2 independent samples • General formula for t statistic – t = (sample statistic –population parameter)/ standard error – Recall APA uses M to designate sample mean • Single sample t – t = (M - )/standard error • Independent means t test – t = ((M1 – M2) – (1 - 2))/standard error Hypothesis test with 2 independent samples • Standard error in t statistic – Measures how accurately the sample statistic measures the population parameter – In the case of a single sample t statistic, the standard error sM represents the error between a sample mean M, and the population mean – In the two sample t statistic, standard error represents the error between the sample mean difference (M1 – M2) and the population mean difference (1 - 2) Hypothesis test with 2 independent samples • Standard error in t statistic – Recall each mean estimate, M1 and M2, has associated with it an error, which can be estimated by the standard error of the mean, – sM = √s2/n (formula for single sample standard error) – Note: errors add even though you are subtracting the two sample means. This makes sense because there are now two sources of error – This means that sM1-M2 = √(s12/n1+ s22/n2) – Caveat: this formula is appropriate only when n1 = n2 Hypothesis test with 2 independent samples • Standard error in t statistic – When n1 ≠ n2 one must used the pooled estimate of the variance – Recall s2 = SS/ df – Pooled estimate of variance = Sp2 = (SS1 + SS2)/(df1 + df2) – Note: Sp2 is the average of the 2 variances. Hence it must be between the two variance estimates Hypothesis test with 2 independent samples • Standard error in t statistic – Two sample standard error = sM1-M2 – sM1-M2 = √(sp2/n1+ sp2/n2) • t statistic for independent measures – t = ((M1 – M2) – (1 - 2))/ sM1-M2 • df for the t statistic = df1 + df2 = n1- 1 + n2 -1 Hypothesis test with 2 independent samples Sample Pop’n Std error data parameter t single M √(s2/n sample t indepen. M1- M2 1 - 2 √(sp2/n1+ sp2/n2) measures Hypothesis test with 2 independent samples • Single sample variance – S2 = SS/df • Two independence sample variance – Sp2 = (SS1 + SS2)/(df1 + df2) Hypothesis test with 2 independent samples • Hypothesis test with independent measures t statistic – See related example 10.1 • Purpose: to determine whether method of loci instructions improve memory for concrete nouns. Design: Between subjects design: group 1 (no specific instructions), group 2 (method of loci); n= 10 participants in each group Hypothesis test with 2 independent samples • Hypothesis – H0 : 1 - 2 = 0 memory performance does not differ between the two treatment conditions – H1 : 1 - 2 ≠ 0 imagery makes a difference • Locate critical region – Set alpha = .05 – Df = df1 + df2 = 10 – 1 + 10 – 1 = 18 – Look up t (18) value ± 2.101 Hypothesis test with 2 independent samples • Calculate test statistic – Results – Group 1 (no imagery): n = 10, M = 19, SS = 40 – Group 2 (imagery): n = 10, M = 26, SS = 50 – Sp2 = (SS1 + SS2)/(df1 + df2) = (40 + 50)/ (9+9) – = 90/18 = 5.0 – sM1-M2 = √(sp2/n1+ sp2/n2) = √(5/10 + 5/10 = 1 Hypothesis test with 2 independent samples • Calculate test statistic – t = ((M1 – M2) – (1 - 2))/ sM1-M2 = (19 – 26) / 1 = -7 • Make decision – M1 = 19; SD = √SS/df = √40/9 = √4.44 = 2.11 – M2 = 26; SD = √SS/df = √50/9 = √5.55 = 2. 36 – t (18) = -7 , p < .05, two tailed. Data suggest that memory performance under imagery conditions is higher than under no imagery instructions Hypothesis test with 2 independent samples • Using descriptive statistics to check plausibility of inferential conclusion – M1 = 19; SD = √SS/df = √40/9 = √4.44 = 2.11 – M2 = 26; SD = √SS/df = √50/9 = √5.55 = 2. 36 What these data are saying is that most of the scores associated with Group 1 fall between 17 and 21, whereas most of the scores associated with Group 2 fall between 23-24 and 28-29. Since these distributions barely overlap, it is plausible to conclude that the two means differ significantly from each other Recall in z distribution ± 1 SD contains about two thirds of the data Hypothesis test with 2 independent samples • Using descriptive statistics to check plausibility of inferential conclusion – M1 = 19; SD = √SS/df = √40/9 = √4.44 = 2.11 – M2 = 26; SD = √SS/df = √50/9 = √5.55 = 2. 36 • Note: in order to check plausibility of conclusions it is important to take into consideration – Magnitude of difference between the means – The size of the SD – The sample sizes of the two distributions Hypothesis test with 2 independent samples • Assumptions – Observations are independent – Two populations must be normally distributed – Two populations must have equal variances Hypothesis test with 2 independent samples • Assumptions – The first two assumptions were already discussed in the one-sample t test – The third assumption is referred to as homogeneity of variance – Although modest violations of homogeneity do not affect interpretation of data, more severe violations make interpretation difficult Hypothesis test with 2 independent samples • Testing for homogeneity of variance (Hartley’s F-max statistic) – Rationale: if population variances are homogeneous, then sample variances should also be similar – Thus, ratio of the two sample variances should be close to 1 Hypothesis test with 2 independent samples • Testing for homogeneity of variance (Hartley’s F-max statistic) – Procedure: compute sample variance for each sample: recall s2 = SS/ df – Select largest and smallest s2 and compute ratio F-max = s2 (largest)/ s2 (smallest) Note: f-max should be close to 1 if there is homogeneity; compare value of F-max to Table B.

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posted: | 7/4/2012 |

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