You are the Director of a large investment fund in Canada. One of your top priorities is to expand your team of mutual fund managers. You are reviewing the resumes of two candidates, Laure Marshall...

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Answered Same DayNov 20, 2021

Answer To: You are the Director of a large investment fund in Canada. One of your top priorities is to expand...

Rajeswari answered on Nov 21 2021
132 Votes
1 (a) It is well-known that mutual funds, on average, underperform their market benchmark. That is, across all mutual funds and many years, the proportion of funds that beat their market benchmark in terms of annual performance is 0.475. You have a strong reason to believe that both Laure and Kenneth are better than average fund managers, otherwise the HR coordinator would not have forwarded you their CV in the first place. Using hypothesis testing, examine whether your prior belief about their abilities is accurate.
(i) H0: x bar = 0.475 vs Ha: x bar <0.475 for both Laure and Kenneth
(ii) Descriptive statistics is as follow:
    Kenneth
     
    Laure
     
     
     
     

     
    Mean
    0.7
    Mean
    0.866667
    Standard Error
    0.15275252
    Standard Error
    0.090851
    Median
    1
    Median
    1
    Mode
    1
    Mode
    1
    Standard Deviation
    0.48304589
    Standard Deviation
    0.351866
    Sample Variance
    0.23333333
    Sample Variance
    0.12381
    Kurtosis
    -1.2244898
    Kurtosis
    4.349112
    Skewness
    -1.0350983
    Skewness
    -2.40476
    Range
    1
    Range
    1
    Minimum
    0
    Minimum
    0
    Maximum
    1
    Maximum
    1
    Sum
    7
    Sum
    13
    Count
    10
    Count
    15
    Confidence Level(95.0%)
    0.34555021
    Confidence Level(95.0%)
    0.194857
    conf interval 95%
    
    
    
    
    Kenneth
    0.35445
    1.04555
    Laure
    0.67181
    1.061523
    
    
    
    
    
    
    
(iii)
    Conclusion: Confidence interval even lower limit is
    only for Laure. Keneeth lower limit is very much
    below 0.425
    
    
    But for both 0.425 is contained in confidence interval
    Also mean is very much higher than 0.425
Hence we can state that Both Kenneth and Laure are performing well above average fixed.
1 (b) Based on the past performance of Laure and Kenneth as fund managers, can you confidently say that Laure is a better fund manager than Kenneth? Does your answer depend on whether you run a 1-tailed test or a 2-tailed test?
(i) Clearly state your null and alternative hypotheses for both tests.
To compare Laure and Kenneth we assume y bar for Laure and x bar for Kenneth.
H0: x bar = y bar vs Ha: x bar (one tailed test for comparison of two means)
(ii) Show your StatTools output for both tests.
    t-Test: Two-Sample Assuming Unequal Variances
     
     
     
     
     
     
    Kenneth
    Laure
    Mean
    0.7
    0.866667
    Variance
    0.23333333
    0.12381
    Observations
    10
    15
    Hypothesized Mean Difference
    0
     
    df
    15
     
    t Stat
    -0.9377617
     
    P(T<=t) one-tail
    0.18161093
     
    t Critical one-tail
    1.75305033
     
    P(T<=t) two-tail
    0.36322187
     
    t Critical two-tail
    2.13144954
     
We find p value is 0.181 which says that there is not significant difference between the mean performances at 5% significant level.
(iii) Briefly state and explain your findings for both tests, making sure to connect them to the relevant portions of the StatTools output.
We find that both are above average estimated as 0.425. Comparing one with other at 5% level there is no significant difference between the performances.
1 (c)  Now reconsider part (b), but suppose that Laure and Kenneth each have twice as much experience (Laure has 30 years, and Kenneth 20 years), and the same success rate as before (13 out of 15 years for Laure, and 7 out of 10 years for Kenneth). See the data under column Q1c. Can you now confidently say that Laure is a better fund manager than Kenneth? Does your answer depend on whether you run a 1-tailed test or a 2-tailed test?
(i) Clearly state your null and alternative hypotheses for both tests.
To compare Laure and Kenneth we assume y bar for Laure and x bar for Kenneth.
H0: x bar = y bar vs Ha: x bar (one tailed test for comparison of two means)
(ii) Show your StatTools output for both tests.
    t-Test: Two-Sample Assuming Unequal Variances
     
     
     
     
     
     
    Kenneth
    Laura
    Mean
    0.7
    0.866667
    Variance
    0.221053
    0.11954
    Observations
    20
    30
    Hypothesized Mean Difference
    0
     
    df
    32
     
    t Stat
    -1.35914
     
    P(T<=t) one-tail
    0.091803
     
    t Critical one-tail
    1.693889
     
    P(T<=t) two-tail
    0.183605
     
    t Critical two-tail
    2.036933
     
(iii) Briefly state and explain your findings for both tests, making sure to connect them to the relevant portions of the StatTools output. Comment specifically on what changes relative to part (b) and why.
Here p value one tailed is reduced to 0.0918 but still larger than 5% our significant level. Thus our conclusion that there is not much significant difference in both performances remain the same but p value can be seen reduced.
2 (a) Run a regression of price on the following variables: year, ppm, colordum, dpi, hp, ibm, apple, and brother. (The printer brand variables are dummy variables, i.e., ibm = 1 if firmname = “IBM/LEXMARK” and 0 otherwise, apple = 1 if firmname = “APPLE_COMPUTER_CO” and 0 otherwise, and brother = 1 if firmname = “BROTHER_INTERNATIONAL_CORP” and 0 otherwise.) see excel file
Copy your StatTools regression output below (make sure your output copies properly and is easy to read).
    SUMMARY OUTPUT
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    Regression Statistics
    
    
    
    
    
    
    Multiple R
    0.771473
    
    
    
    
    
    
    R Square
    0.59517
    
    
    
    
    
    
    Adjusted R Square
    0.588012
    
    
    
    
    
    
    Standard Error
    1093.617
    
    
    
    
    
    
    Observations
    519
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    ANOVA
    
    
    
    
    
    
    
     
    df
    SS
    MS
    F
    Significance F
    
    
    Regression
    9
    8.95E+08
    99443079
    83.14649
    4.3E-94
    
    
    Residual
    509
    6.09E+08
    1195999
    
    
    
    
    Total
    518
    1.5E+09
     
     
     
    
    
    
    
    
    
    
    
    
    
     
    Coefficients
    Standard Error
    t Stat
    P-value
    Lower 95%
    Upper 95%
    Lower 95.0%
    Intercept
    730702.5
    51826.19
    14.0991
    2.41E-38
    628882.9
    832522
    628882.9
    year
    -366.704
    26.05226
    -14.0757
    3.06E-38
    -417.888
    -315.521
    -417.888
    ppm
    248.6547
    15.38393
    16.16327
    9.85E-48
    218.4308
    278.8785
    218.4308
    colordum
    5314.661
    370.1349
    14.35872
    1.69E-39
    4587.481
    6041.842
    4587.481
    dpi
    -0.35301
    0.372857
    -0.94678
    0.344199
    -1.08554
    0.379513
    -1.08554
    age
    205.3629
    43.89311
    4.678706
    3.71E-06
    119.129
    291.5969
    119.129
    ibm
    52.27552
    153.3206
    0.340956
    0.733278
    -248.944
    353.4947
    -248.944
    hp
    246.608
    141.3297
    1.744913
    0.081604
    -31.0533
    524.2693
    -31.0533
    apple
    531.3171
    225.1429
    2.359911
    0.018656
    88.99347
    973.6408
    88.99347
    brother
    -410.952
    201.3021
    -2.04147
    0.041719
    -806.437
    -15.4668
    -806.437
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    RESIDUAL OUTPUT
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    Observation
    Predicted...
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