Homework 07 - Answers and Points

# Homework 07 - Answers and Points -  ...

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Unformatted text preview:   Homework Assignment 7  Psychology 60  Spring 2010    ‐ For homework problems requiring calculation, you must show all of your work,  including intermediate steps, in order to receive credit.        1) (4 points total) A sample of n = 16 individuals participates in a repeated‐measures study that  produces a sample mean difference of  X D = 9  with SS = 960.   a. (1 point) Calculate the variance for the sample of difference scores.  SS 960 2 2 2 sD = D , sD = , sD = 64   df 15 b. (1 point) Briefly explain what the measured by the variance of difference scores  The variance of the difference scores is a measure of the consistency of the effect.     c. (1 point) Calculate the estimated standard error for the sample mean difference  sXD = 2 sD 64 , sXD = , sXD = 4 , sXD = 2   n 16   d. (1 point) Briefly explain what is measured by the estimated standard error in this  situation  The estimated standard error here measures the standard deviation of a sampling  distribution of means of samples of 16 difference scores. That is, it measures typical  amount that a mean of a sample of 16 difference scores will differ from the population  mean of difference scores      2) (9 points total) The following data are from a repeated‐measures study examining the effect of  a treatment by measuring a group of n = 4 participants before and after they receive treatment.    Participant    Before  After    A      7    10    B      6    13    C      9    12    D      5    8    a. (2 points) Calculate the difference scores and  X D   (Either way of subtracting is fine)  Participant    Before  After    Difference Score (Before – After)    A      7    10    ‐3    B      6    13    ‐7    C      9    12    ‐3    D      5    8    ‐3      XD = ΣX D -16 = = -4   n 4   b. (2 points) Compute the SS for the difference scores    ( X Di − X D )     ( X Di − X D )2     Difference Score        ‐3      ‐3 ‐ ‐4 = 1      1     ‐7      ‐7 ‐ ‐4 = ‐3      9     ‐3      ‐3 ‐ ‐4 = 1      1     ‐3      ‐3 ‐ ‐4 = 1      1      SSD = 12     c. (1 point) Compute the sample variance for the difference scores    SS 12 2 2 2 sD = D , sD = , sD = 4   df 3   d. (1 point) Compute the estimated standard error of the difference scores    sXD = 2 sD , sXD = n 4 , s = 1  4 XD   e. (1 point) Compute  t XD     t XD = XD -4 , t XD = , t XD = -4   sXD 1   f. Is there a statistically significant treatment effect? Use a non‐directional hypothesis test  with  α = 0.05     (1 point)  df = 3,  α = 0.05 ,  t critical = ± 3.182      (1 point) |tobserved| > |tcritical|,  |­4| > 3.182, Reject Null Hypothesis    3) (13 points total) A researcher uses a matched‐samples design to investigate whether single  people who own pets are generally happier than single people without pets. A mood inventory  questionnaire is administered to a group of 20‐ to 29‐year old non‐pet owners and a similar age  group of pet owners. The pet owners are matched one‐to‐one with the non‐pet owners for  income,number of close friendships and general health. The data are as follows.    Matched Pair  Non­Pet  Pet       A      11    13    B      9    8    C      11    14    D      13    13    E      6    12    F      9    11    a. Is there a statistically significant difference in the mood scores for non‐pet owners versus  pet owners? Use a non‐directional hypothesis test with  α = 0.05       (2 points) For finding difference scores and mean        (Was listed as Before / After in original HW, sorry)  Pet  13  8  14  13  12  11                Difference (NP – Pet)  ‐2  1  ‐3  0  ‐6  ‐2  Difference (NP – Pet)    ( X Di − X D )       ( X Di − X D )2                           Matched Pair  A      B      C      D      E      F      Non­Pet  11    9    11    13    6    9    ‐2  1  ‐3  0  ‐6  ‐2  ‐2 ‐ ‐2 = 0  1 ‐ ‐2 = 3  ‐3 ‐ ‐2 = ‐1  0 ‐ ‐2 = 2  ‐6 ‐ ‐2 = ‐4  ‐2 ‐ ‐2 = 0                          0  9  1  4  16  0    SSD = 30  ΣX D −12 = = -2   n 6 XD =     (2 points) for finding SS                (1 point)  2 sD =                         SSD 30 2 2 , sD = , sD = 6   df 5   (1 point)     sXD = 2 sD , sXD = n 6 , s = 1  6 XD (1 point)     (1 point)      (1 point)     t XD = XD -2 , t XD = , t XD = -2   sXD 1 df = 5,  α = 0.05 ,  t critical = ± 2.571   |tobserved| < |tcritical|,   |­2| < 2.571, Fail to Reject Null Hypothesis    b. (2 points) Write a sentence describing the outcome of the hypothesis test as it would  appear in a research report    A study was run to investigate whether single people who own pets are generally happier  than single people without pets. No evidence for a difference in happiness was found,  t(5)=‐2, p > 0.05.     c. (2 points) Why does the matching of subjects in this way increase the power of this test?    Individual differences in happiness that are attributable to income, number of close  friendships and general health are removed from the analysis as a consequence of the  matching.     4) (2 points)  A repeated‐measures study and an independent‐measures study both produced a t  statistic with df = 16.  How many individuals participated in each study?    For the repeated measures study, df = (nD ‐1), thus there were 17 difference scores, which  means 17 individuals participated.     For the independent measures study, df = (n1‐1) + (n2 ‐1), or (ntotal – 2), thus there were 18  individuals total.     5) (2 points) A matched‐subjects study and an independent‐measures study both produced a t  statistic with df = 16.  How many individuals participated in each study?    For the matched subjects study, df = (nD ‐ 1), thus there were 17 difference scores; difference  scores are between different individuals, thus there were 34 participants.    For the independent measures study, df = (n1‐1) + (n2 ‐1), or (ntotal – 2), thus there were 18  individuals total.    6) (1 point) A researcher conducts a research study comparing two treatment conditions and  obtains 30 scores in each treatment.  If the researcher used a repeated‐measures design, then  how many individuals participated in the research study?    Repeated measures studies have measures for each subject twice, thus 30 people  participated in this study, giving 60 total scores, 30 in each treatment.       7) (2 points)  Briefly explain the advantages and disadvantages of using a repeated‐measures  design as opposed to an independent‐measures design.    The main advantage to a repeated measures design is that variability due to individual  differences is removed prior to analysis because individuals are compared to themselves.  The main disadvantage is carry‐over and order effects; that is, the effect of one treatment can  leak into another, and that one treatment always comes before or after another.  In  independent measures designs one does not have to worry about order and carry over  effects, but variability due to individual differences is measured together with the effect of  some treatment or manipulation, leading to less power to detect the effect.     ...
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