PSYC2005 (RDA IIA) Exam I (Statistics) October/November 2011

PSYC2005 (RDA IIA) Exam I (Statistics) October/November 2011   Question 1 a)      Correlation (use formula for Pearson’s r). The answer is r = + 0.1229 b)      Very weak, positive linear relationship – although yearmark and exam mark move in the same direction, they are only very weakly related.   Question 2 i)        Comparison of two groups (librarians and designers) – no reason to link a specific librarian to a specific designer therefore two independent groups. Assume parametric therefore run a t-test for two independent samples. ii)       Null hypothesis would be H0: Mu 1 = Mu 2 (or Mu p = Mu q) iii)     Alternate hypothesis would be H1: Mu 1 not equal to Mu 2 (or Mu p not equal to Mu 2) because told to test for a difference iv)     Alpha = 0.05 v)      n is 10 for group 1 and 10 for group 2, the degree of freedom is therefore 10 plus 10 minus 2 which is 18.  The test statistic would be t18 = 1.4574. vi)     The critical areas would be 0.025 in each tail; the corresponding critical values from t-tables would be 2.101 and -2.101. vii)   1.4574 is outside the critical areas therefore Fail to Reject H0 at alpha = 0.05. viii)  There is insufficient evidence at the 0.05 level of significance to believe that there is a significant difference between the librarians and designers in mean time to recognize the word.   Question 3 Sampling distribution: Mean IQ (x-bar) is distributed normally with a mean of 100 and a variance of 16-squared over 36 (standard deviation of 16 over the square root of 36). The probability that the average IQ is at least 102 (i.e. greater than 102) is 0.2266 (z =  0.75).   Question 4 i)        Comparison of three groups, parametric therefore ANOVA ii)       Null hypothesis would be H0: Mu 1 equal to Mu 2 = Mu 3 (all the means are equal) iii)     Alternate hypothesis would be H1: at least one pair of means not equal iv)     Alpha equals 0.05 v)      SS error  = 511.46, k = 3, df between = 2, N = 30, df error = 27, df total = 29, MS between = 90.535, MS error = 18.943, test statistic is F (2; 27) = 4.7793 vi)     The critical area would be 0.05 in the right tail; the corresponding critical value from F-tables would be 3.35.  vii)   4.7793 is inside the critical area therefore Reject H0 at alpha = 0.05. viii)  There is sufficient evidence at the 0.05 level of significance to believe that there is a significant difference between the time to complete the task between at least two of the conditions   Question 5 i)        Association between two nominal variables (gender – male/ female and driving speed – below limit, at limit, above limit) therefore Chi-squared test of association  ii)       Null hypothesis would be H0: there is no relationship between gender and driving speed iii)     Alternate hypothesis would be H1: there is a relationship between gender and driving speed  iv)     Alpha equals 0.05 v)      The test statistic would be Chi-squared 2 = 35.796 (please refer to hand out for steps). vi)     The critical area would be 0.05 in the right tail but please use the non-directional row on the table; the corresponding critical value from Chi-squared -tables would be 5.99.  vii)   35.796 is inside the critical area therefore Reject H0 at alpha = 0.05. viii)  There is sufficient evidence at the 0.05 level of significance to believe that there is a relationship between gender and driving speed   Question 6   a)      IV = dominance = nominal; DV = span in centimetres = ratio. Comparison of two linked measurements (groups) therefore matched. DV is at least interval, no other information given re other parametric assumptions (although the sample size is 15 which is too small for Central Limit Theorem to apply and could indicate that the data might not be normally distributed). If assume the other parametric assumptions, including normality, are met, could run a parametric matched pairs t-test; if assume the other assumptions, for example, normality, are not met, would run a Wilcoxan’s MPSR test.  b)      IV = background risk factors composite score – at least ordinal (probably interval); DV = substance abuse score – at least ordinal (probably interval). Assessing relationship therefore use correlation/ regression. c)       IV = TV programme type = nominal; DV = serotonin level  measurement = interval (could argue ratio). Comparison of four groups, DV is at least interval, no other information re parametric assumptions is given. Assuming these are met, could run a parametric ANOVA, if not met could run a Kruskal-Wallis. d)      IV = reading ability = nominal; DV = IQ = interval. One group/sample compared to a population where the standard deviation/ variance for the population is known, therefore use a z-test e)      IV = TV viewership = nominal; DV = annual income = nominal. Relationship between two nominal variables therefore use a Chi-squared test of association.     Question 7 a)      False, zero represents the absence of measurement on a ratio scale. b)      False, a statistic can take on different values and is therefore not fixed. c)       False, the dependent variable in a regression line can be accurately estimated using any value that is within the range of recorded values for the independent variable. Outside this range is extrapolation. d)      True, sample size is n and the standard deviation of the sampling distribution is sigma over the square root of n. As n gets larger, one would divide by a larger number and the answer would get smaller. e)      False, the null hypothesis assumes the status quo i.e. no change and the alternate hypothesis proposes what the experimenter would like to prove.