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    Chi square test of independence, Confidence Interval

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    Problem 23: In one part of a test developed by a psychologist, the test subject is asked to form a word by unscrambling the letters "ciiatttsss". Given below are times (in seconds) required by 15 randomly selected persons to unscramble the letters. Construct a 95% confidence interval for standard deviation. Assume the population is approximately normal.

    Times (in seconds)
    68.7
    27.4
    26
    60.5
    34.6
    61.1
    68.6
    48.4
    43.6
    39.5
    85.3
    26.3
    43.4
    83.7
    68.9

    Problem 24: Doing the "Statistics Modules" seems to help statistics students on their exams. Students in several statistics class are classified by grade in the course and by how many modules they completed. There wer 18 possible modules. The results are listed in the following table.

    Grade in course 16 or more 9 to 15 8 or less
    A 25 9 3
    B 18 11 6
    C 5 14 12
    D or F 2 10 14

    Does the data suggest that course grade and number of modules completed are dependent? Use an alpha=0.05 level of significance

    © BrainMass Inc. brainmass.com October 2, 2022, 5:39 pm ad1c9bdddf
    https://brainmass.com/statistics/chi-squared-test/chi-square-test-independence-confidence-interval-6383

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    23 In one part of a test developed by a psychologist, the test subject is asked to form a word by unscrambling the letters "ciiatttsss". Given below are times (in seconds) required by 15 randomly selected persons to unscramble the letters. Construct a 95% confidence interval for standard deviation. Assume the population is approximately normal.

    Times (in seconds)
    68.7
    27.4
    26
    60.5
    34.6
    61.1
    68.6
    48.4
    43.6
    39.5
    85.3
    26.3
    43.4
    83.7
    68.9

    First we calculate variance and standard deviation of the data
    Score (Score-mean)^2
    68.7 265.69000
    27.4 625.00000
    26 696.96000
    60.5 65.61000
    34.6 316.84000
    61.1 75.69000
    68.6 262.44000
    48.4 16.00000
    43.6 77.44000
    39.5 166.41000
    85.3 1082.41000
    26.3 681.21000
    43.4 81.00000
    83.7 979.69000
    68.9 272.25000
    786.00 5664.64000
    n= 15
    mean= 52.4 =786/15
    Variance=s 2 404.6171429 =5664.64/14
    standard deviation= 20.1151 =square root of 404.617143

    Find Chi square values for 95% confidence interval
    Degrees of fredom=15-1=14
    chi square vlaues
    X upper ^2 = 26.11894805
    X lower ^2 = 5.628726168
    Variance Standard deviation=square root of Variance
    s L 2 = (n-1) s ^2 / X upper ^2 = 216.87857 14.7268
    s U 2 = (n-1) s ^2 / X lower ^2 = 1006.38045 31.7235

    95% confidence interval
    Lower limit of standard deviation 31.7235
    Upper limit of standard deviation 14.7268

    Problem 24

    Doing the "Statistics Modules" seems to help statistics students on their exams. Students in several statistics class are classified by grade in the course and by how many modules they completed. There wer 18 possible modules. The results are listed in the following table.

    Grade in course 16 or more 9 to 15 8 or less
    A 25 9 3
    B 18 11 6
    C 5 14 12
    D or F 2 10 14

    Does the data suggest that course grade and number of modules completed are dependent? Use an alpha=0.05 level of significance

    Null Hypothesis: Course grade and number of modules completed are independent
    Alternative Hypothesis: Course grade and number of modules completed are dependent

    Theoretical=Row total X column total/n
    Thus for Grade A and >16 theoretical=37*50/129= 14.3411
    and so on

    No of modules completed

    >16 9-15 <8
    Grade in course Actual Theoretical Actual Theoretical Actual Theoretical Row total

    A 25 14.3411 9 12.6202 3 10.0388 37

    B 18 13.5659 11 11.938 6 9.4961 35

    C 5 12.0155 14 10.5736 12 8.4109 31

    D or F 2 10.0775 10 8.8682 14 7.0543 26
    129

    Column total 50 44 35

    Chi square value= summation of (fo-fe)^2/fe= 36.9017

    fo fe (fo-fe)^2/fe
    25 14.3411 7.92214

    18 13.5659 1.44931

    5 12.0155 4.09615

    2 10.0775 6.47442

    9 12.6202 1.03848

    11 11.938 0.0737

    14 10.5736 1.11033

    10 8.8682 0.14445

    3 10.0388 4.93532

    6 9.4961 1.28713

    12 8.4109 1.53154

    14 7.0543 6.83877
    36.9017

    No of degrees of freedom= (No of rows-1)*(No of columns-1)=(4-1)*(3-1)=6

    alpha (a) = 0.05

    Chi square value= 12.5916 from the tables

    Accptance region for null hypothesis is upto 12.5916

    Since Chi square value computed in this case= 36.9017
    which lies to the right of acceptance region of 12.5916
    Reject Null Hypothesis
    Course grade and number of modules completed are dependent

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com October 2, 2022, 5:39 pm ad1c9bdddf>
    https://brainmass.com/statistics/chi-squared-test/chi-square-test-independence-confidence-interval-6383

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