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There are sixty (60) multiple-choice questions in this section. Answer all questions. Select the single best response and record the appropriate letter, e.g. A. For each question, write your answer in...

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There are sixty (60) multiple-choice questions in this section. Answer all questions. Select the single best response and record the appropriate letter, e.g. A. For each question, write your answer in the answer sheet(s) provided. DO NOT circle your answers on the question paper.
Answered Same Day Apr 19, 2020

Solution

Abr Writing answered on Apr 21 2020
139 Votes

Sheet1
        Part 1                                    Ques no.    Option
        Year    2013    2014    2015    2016    2017                1    B
        Return    5%    -1%    2%    4%    7%                2    D
                                            3    C
        mean    3.40%                                4    C
        Mode    NA                                5    D
        Median    4.00%                                6    C
                                            7    B
        Part 2                                    8    B
            Large cap stocks     Small cap stocks                            9    D
        Year 1    16%    14%            part 20                10    A
        Year 2    -10%    -8%            Std e
or    std dev/sqrt(n)            11    D
        Year 3    12%    15%                0.25%            12    B
                                            13    A
        Std dev    14.0%    13.0%            part 21                14    C
                            Std e
or    std dev/sqrt(n)            15    B
        Part 3                        0.1875%            16    B
        Year    1    2    3        Part 26                17    A
        Ret    8%    10%    7%        Std e
or    0.0277180765            18    C
                            UL    0.2127180765            19    D
        Geometric mean    0.082425706                LL    0.1572819235            20    B
                                            21    A
        part11    P(s)                Part 28                22    B
            0.2    mean = n*p = 0.2*5 =1            Std e
or    0.0059950216            23    D
            0.2    var = n*p*(1-p) = 0.8            UL    6.600%            24    A
            0.2                LL    5.40%            25    C
            0.2                                26    D
            0.2                Paart 30                27    A
                            alpha/2    0.005            28    B
        Part 12                    Z    2.5758293035            29    B
        4C2*P^2*(1-p)^2                    Size    200            30    A
        0.5248                    p    36%            31    D
                            SE    0.0339411255            32    A
                            Margin of e
or    0.0874265457            33    B
        Part 13                    UL    44.743%            34    C
        S    58-49+1                LL    27.257%            35    B
        T    100-0+                                36    C
                            Paart 31                37    C
        P    S/T    0.1            alpha/2    0.025            38    D
                            Z    1.9599639845            39    D
        Part 14                    Size    150            40    C
        0.6914624613                    p    22.00%            41    B
                            SE    0.0338230691            42    C
        Part 15                    Margin of e
or    0.0662919972            43    A
        0.2511465536                    UL    28.629%            44    B
                            LL    15.371%            45    B
        Part 16                                    46    C
        p(d)    0.2                Part 37                47    A
        Mean    n*p    0.8            x    350000            48    A
        Variance    n*p*(1-p)    0.64            x bar    300000            49    C
                            std dev    150000            50    A
                            Test stat    2.00            51    A
        Part 17                    Critic Val    2.326347874            52    D
        10C1(0.04)^1*(0.96)^9+10c0*(0.96)^10                                    53    D
        0.9418462343                    Part 40                54    D
                            x    250            55    B
                            x bar    260            56    C
                            std dev    75            57    C
                            Test stat    -0.73            58    B
                            Critic Val    2.326347874            59    B
                                            60    D
                            Part 42
                            The z-statistic is computed as follows:        0.2436
                            z = \frac{\bar p - p_0}{\sqrt{p_0(1-p_0)/n}} = \frac{ 0.46 - 0.42 }{\sqrt{ 0.42(1- 0.42)/100}} = 0.81z=p0​(1−p0​)/n​p¯​−p0​​=0.42(1−0.42)/100​0.46−0.42​=0.81    0.81    0.002436
                                    0.0493558507
                                    0.8104408985
                            Part 45
                             Sample 1      Sample 2     Difference 
                            Sample proportion    0.38    0.3    0.08
                            95% CI (asymptotic)    0.2736 - 0.4864    0.1996 - 0.4004    -0.2936
                            z-value    1.1
                            P-value    0.2855
                            Interpretation    Not significant,
                                accept null hypothesis that
                                sample proportions are equal
                            n by pi    n * pi >5, test ok
                            Part 52
                            Day    SA    SB
                            1    2.05%    -0.08%
                            2    1.13%    2.23%
                            3    0.28%    0.45%
                            4    -0.15%    -3.13%
                            5    2.75%    0.38%
                            First create a table of the variable scores, their ranks, differences in ranks (di) and the differences in ranks squared (di2).
                            Score 1    Score 2    Rank 1    Rank 2    di    di2
                            2.05%    -0.08%    4    1    3    9
                            1.13%    2.23%    3    5    2    4
                            0.28%    0.45%    2    4    2    4
                            -0.15%    -3.13%    1    2    1    1
                            2.75%    0.38%    5    3    2    4
                            You now need to add up (sum) the differences in ranks squared (di2).
                                22
                                22
                            As you have no ties in your data use the following formula:
                            Substitute the sum of di2 information from Step 2 and the number of scores in each variable (n) into the formula:
                                6 x 22
                                5[(5)2 - 1]
                                132                0.08
                                120                7%
                                1.1                0.0056        5.3571428571
                                -0.1
                            Part 53
                            Co
(x,y)    Covar(x.y)/stddev(x)*stddev(y)
                            Co
(x,y)    0.4441882553
                            Part 57
                            Co
el    0.0535714286
                            Wa    45%
                            Wb    55%
                            Std dev a    8%
                            Std dev b    7%
                            Var    0.29%
61-80
        Part 61-65                                Part    Ans
        Month     Excess return (Xt)                            61    D
        Jan    -0.02%                            62    A
        Feb    -0.10%                            63    A
        Mar    0.10%                            64    B
        Apr    0.30%                            65    C
        May    -0.50%                            66    B
        Jun    0.60%                            67    A
        Jul    0.20%                            68    D
        Aug    0.90%                            69    C
        Sep    -0.20%                            70    B
        Oct    0.40%                            71    D
        Nov    -0.10%                            72    B
        Dec    0.50%                            73    C
        Mean    0.17333%                            74    C
        Std dev    0.0038962297                            75    B
        SE    0.0011247446                            76    A
        The provided sample mean is \bar X = 0.001733X¯=0.001733 and the sample standard deviation is s = 0.00389623s=0.00389623, and the sample size is n = 12n=12.                                77    C
        (1) Null and Alternative Hypotheses                                78    C
        The following null and alternative hypotheses need to be tested:                                79    B
        Ho: \muμ = 00                                80    B
        Ha: \muμ > 00
        This co
esponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.
        (2) Rejection Region
        Based on the information provided, the significance level is \alpha = 0.05α=0.05, and the critical value for a right-tailed test is t_c = 1.796tc​=1.796.
        The rejection region for this right-tailed test is R = {t: t > 1.796}R=t:t>1.796
        (3) Test Statistics
        The t-statistic is computed as follows:
        t = =1.541
        (4) Decision about the null hypothesis
        Since it is observed that t = 1.541 \le t_c = 1.796t=1.541≤tc​=1.796, it is then concluded that the null hypothesis is not rejected.
        Using the P-value approach: The p-value is p = 0.0758p=0.0758, and since p = 0.0758 \ge 0.05p=0.0758≥0.05, it is concluded that the null hypothesis is not rejected.
        (5) Conclusion
        It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean \muμ is greater than 0, at the 0.05 significance level.
        Confidence Interval
        The 95% confidence interval is -0.001 < \mu < 0.004−0.001<μ<0.004.
        Part 66-70
        year    Quarter    Manager    Fund    Excess
        2015    1    15%    14%    1.00%
        2015    2    6.10%    5.10%    1.00%
        2015    3    12%    11.05%    0.95%
        2015    4    16.20%    14.10%    2.10%
        2016    1    7.35%    6.35%    1.00%
        2016    2    6.80%    5.70%    1.10%
        2016    3    11.25%    10.05%    1.20%
        2016    4    11.25%    10.25%    1.00%
        2017    1    0.50%    0%    0.50%
        2017    2    3%    2.05%    0.95%
        2017    3    14%    11.65%    2.35%
        2017    4    9%    7%    1.80%
            Mean    9.35%    8.11%    1.2458%
            Std dev    4.81%    4.49%    0.5437%
            Std e
or    1.39%    1.30%    0.1570%
        The provided sample mean is \bar X = 0.012458X¯=0.012458 and the sample standard deviation is s = 0.005437s=0.005437, and the sample size is n = 12n=12.
        (1) Null and Alternative Hypotheses
        The following null and alternative hypotheses need to be tested:
        Ho: \muμ = 00
        Ha: \muμ > 00
        This co
esponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.
        (2) Rejection Region
        Based on the information provided, the significance level is \alpha = 0.01α=0.01, and the critical value for a right-tailed test is t_c = 3.106 tc​=3.106
        The rejection region for this right-tailed test is R = {t: t > 2.718}R=t:t>2.718
        (3) Test Statistics
        The t-statistic is computed as follows:
        t =7.937
        (4) Decision about the null hypothesis
        Since it is observed that t = 7.937 > t_c = 2.718t=7.937>tc​=2.718, it is then concluded that the null hypothesis is rejected.
        Using the P-value approach: The p-value is p = 0p=0, and since p = 0 < 0.01p=0<0.01, it is concluded that the null hypothesis is rejected.
        (5) Conclusion
        It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean \muμ is greater than 0, at the 0.01 significance level.
        Confidence Interval
        The 99% confidence interval is 0.008 < \mu < 0.0170.008<μ<0.017.
        Part 71-75
        Month    Stock 1    Stock2
        1    2.75%    -0.03%
        2    0.79%    2.30%
        3    1.78%    2.73%
        4    1.67%    -1.73%
        5    -2.09%    -2.65%
        6    3.53%    1.57%
        7    1.55%    -3.68%
        8    3.67%    -0.27%
        9    3.72%    3.11%
        10    2.96%    5.91%
        11    2.35%    -3.76%
        12    -1.91%    4.97%
        Mean
        t-Test: Paired Two Sample for Means
            Variable 1    Variable 2
        Mean    0.0173083333    0.0070583333
        Variance    0.0003878681    0.0010572699
        Std dev    0.0196943674    0.0325156874
        Observations    12    12
        Pearson Co
elation    0.0591192783
        Co var    0.0038%
        df    11
        t Stat    0.9595022801
        P(T<=t) one-tail    0.1789607378
        t Critical one-tail    1.7958848187
        P(T<=t) two-tail    0.3579214756
        t Critical two-tail    2.2009851601
        Part 76-80
        Std dev A    0.0358887169
        Std dev B    0.0532916504
        Covar    0.01%
        Co
el    0.0726771183
            A    B
        Expected ret    1%    1.26%
        x*1% + (1-x)*1.26% = 1.05%
    X    0.8076923077
            WtA    WtB
            65%    35%
        Portfolio std dev    0.0309083322
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