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Data MARK GENDER CLASS STATUS 84.84 1 1 1 GENDER 1 FEMALE 63.64 1 2 1 2 MALE 54.43 1 3 1 55.65 1 3 1 CLASS 1 CLASS 1 66.40 1 2 1 2 CLASS 2 74.51 1 1 1 3 CLASS 3 58.73 1 3 1 43.64 1 3 1...

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Data MARK GENDER CLASS STATUS 84.84 1 1 1 GENDER 1 FEMALE 63.64 1 2 1 2 MALE 54.43 1 3 1 55.65 1 3 1 CLASS 1 CLASS 1 66.40 1 2 1 2 CLASS 2 74.51 1 1 1 3 CLASS 3 58.73 1 3 1 43.64 1 3 1 STATUS 1 DOMESTIC 70.20 1 2 1 2 INTERNATIONAL 73.30 1 2 1 98.23 1 1 1 27.90 1 3 1 33.67 1 3 1 58.37 1 2 1 56.72 1 2 1 76.18 1 1 2 76.27 1 1 2 78.13 1 1 2 45.53 1 3 2 37.06 1 3 2 51.96 1 2 2 26.18 1 3 2 56.73 1 1 2 54.24 1 2 2 51.07 1 2 2 39.82 1 3 2 98.66 1 1 2 91.04 1 1 2 90.11 1 1 2 68.00 1 2 2 42.77 2 3 1 43.98 2 3 1 52.37 2 1 1 44.38 2 3 1 48.91 2 2 1 46.57 2 2 1 67.58 2 1 1 55.50 2 2 1 48.15 2 3 1 38.63 2 2 1 64.82 2 1 1 45.24 2 3 1 54.31 2 1 1 53.38 2 1 1 40.48 2 3 1 50.84 2 1 2 52.15 2 3 2 40.60 2 2 2 63.43 2 1 2 45.92 2 3 2 47.89 2 3 2 43.65 2 2 2 47.12 2 2 2 60.18 2 1 2 42.52 2 3 2 46.70 2 2 2 42.44 2 2 2 49.19 2 2 2 59.71 2 1 2 66.11 2 1 2 STATISTICS FOR DECISION MAKING BUSINESS REPORT XXXXXXXXXXName: Tejas Kaja XXXXXXXXXXStudent no: XXXXXXXXXX XXXXXXXXXXCampus: Wollongong Executive Summary: With (F=27.37, p<0.05), one-way ANOVA along with the Post Hoc analysis concludes that there is a significant difference in the mean marks between (class1, class2), (class1, class3), and (class2, class3). With (t=-0.25, p>0.05), the T-test of independent samples indicates that there is 79.9% probability that there is no significant difference in the mean marks between two types of student status namely domestic and international. It is recommended to give equal priority to the domestic and international type of students regarding faculty and teaching methods. With (Chi(2)=1.6, p>0.05), the Chi square test of Independence indicates that there is 44.9% chance that type of class and type of student status are independent of each other. With (z=0.5346, p>5%), one sample Z test indicates that there is 29.6% chance that the mean marks of students are less or equal to 55. adopting a practical method of teaching, proper invigilation on students regarding their homework an assignment might be helpful to improve the proportion of students whose mean marks are greater than 55. With (F=5.93, p<0.05), the f test for variance indicate that there is a significant difference in the variants marks between male and female students. With (t=3.04, p<0.05), the T-test of independent samples indicates that there is sufficient evidence to conclude that there is a significant difference in the mean marks of male and female students. With (z=3.126, p<5%), the Z test for difference in proportion indicates that proportion of Marks greater than 55 for females is higher than male students. Checking attendance on a regular basis and an introduction of the penalty for missing classes can improve the proportion of marks greater than 55 for male students. Table of Contents Executive Summary 1 Business Problem 3 Statistical Problem 4 Analysis 5 Hypothesis 1 5 Hypothesis 2 8 Hypothesis 3 10 Hypothesis 4 12 Hypothesis 5 13 Hypothesis 6 15 Hypothesis 7 17 Conclusions 19 Implications 20 Business Problem The data for marks of 60 students along with their gender, class, and status is to be analyzed. The main area of concern is to test the proportion of students with their marks less than 55. Comparison of the proportion of male and female students with marks greater than 55 is also an objective. Other objectives include knowing the independence of type of student status and gender. Lastly the test for difference in the mean marks between the type of student as well as the type of gender might be beneficial.   Statistical Problem Various types of parametric hypothesis testing are used to analyze my objective. One way ANOVA is used for testing difference between the mean marks of various classes. T-test for independent samples are helpful for testing the mean difference in marks between two categories of student status as well as between genders.  Chi square test of Independence is used to test the dependence of type of class and the type of student status. Sample Z test seems appropriate for testing if mean marks of students are greater than 55. F test to test the equality of variances of marks between male and female students. Z test for difference in proportions is beneficial for testing if the proportion of Marks greater than 55 for females is higher than male students. Analysis Hypothesis 1 The null hypothesis, ho: the mean marks of various classes namely class 1, class 2 and class 3 don’t differ significantly. The alternative hypothesis, H1: at least one of the mean marks of various classes namely class 1, class 2 and class 3 differs significantly. For the application of one way ANOVA, the dependent variable should be measured by ratio scale of measurement. And the independent variable is categorized into three or more than three groups. In this case, my dependent variable is marks of students. The independent variable is different categories of class (1, 2, and 3). The calculation of F test statistic on the basis of ANOVA is done with the help of following formula. Carlin, B.P. and Louis, T.A., 2010.  source Df SS MSS F between dfb = k - 1 SSB = ∑_j {nj (xbar_j-xbar)2} MSB = SSB / dfb F = MSB / MSW within dfw = n - k SSW = ∑_j∑_i(xij-xbar_j )2   MSW = SSW / dfw   total dft = n - 1 SST = ∑_j∑_i(xij-xbar)2      The output as obtained from Excel for the analysis of one way ANOVA is given below. ANOVA: Single Factor SUMMARY Groups Count Sum Average Variance Class1_marks 20 1417.43 70.8715 XXXXXXXXXX Class2_marks 20 1073.21 53.6605 XXXXXXXXXX Class3_marks 20 876.09 43.8045 70.1333 ANOVA Source of Variation SS df MS F P-value F crit Between Groups XXXXXXXXXX 2 XXXXXXXXXX 27.3753 0.0000 3.1588 Within Groups XXXXXXXXXX 57 XXXXXXXXXX Total XXXXXXXXXX 59 Level of significance 0.05 With (F=27.37, p<0.05), I reject the null hypothesis 5% level of significance. Hence there is enough evidence to support the claim that at least one of the mean marks between three categories of classes differs significantly. To test which pair of classes differ significantly I use post hoc analysis, Turkey Crammer test. The formula for Turkey Crammer test is shown below. Critical Range = QU(c,n-c) * sqrt{(MSW/2) * (1/nj +1/nj')} Where n = total number of values in that group Nj = number of values in jth group c = number of groups A specific pair is said to be significantly different if the absolute mean difference is greater than the critical range. The Excel output of Turkey Crammer analysis is obtained from Excel is given below. Wilcox, R.R., 1996.  Tukey-Kramer Multiple Comparisons Sample Sample Group Mean Size 1: Class1_marks 70.8715 20 2: Class2_marks 53.6605 20 3: Class3_marks 43.8045 20 Other Data Level of significance 0.05 Numerator d.f. 3 Denominator d.f. 57 MSW XXXXXXXXXX Q Statistic 3.403 Absolute Std. Error Critical Comparison Difference of Difference Range Results Group 1 to Group 2 17.211 XXXXXXXXXX 8.9099 Means are different Group 1 to Group 3 27.067 XXXXXXXXXX 8.9099 Means are different Group 2 to Group 3 9.856 XXXXXXXXXX 8.9099 Means are different There is a significant difference in the mean marks between (class1, class2), (class1, class3), and (class2, class3). Box, G.E., Hunter, J.S. and Hunter, W.G., 2005.  Hypothesis 2 The null hypothesis, Ho: the mean marks between two types of student status namely domestic and international don’t differ significantly. The alternative hypothesis, h1: the mean marks between two types of student status namely domestic and international differs significantly. Box, G.E., Hunter, J.S. and Hunter, W.G., 2005.  T-test for independent samples with unequal variances is applied to test this hypothesis. The dependent variable is marks of students. Independent variable is the type of student status which is categorized as domestic and international. Since domestic and International Group is independent of each other, it is sufficient to apply the T-test for independent samples. The test statistic for T-test of independent samples with unequal variances is given below. Wilcox, R.R., 1996.  T = (Xbar1-Xbar2) / √((s12)/n1 +(s22)/n2 ) Degrees of freedom, df = ((s12)/n1 +(s22)/n2 )2 / (1/(n1-1) ((s12)/n1)2+1/(n2-1) ((s22)/n2)2) The Excel generated output for testing the difference in mean marks between two types of student status is given below. Separate-Variances t Test for the Difference Between Two Means (assumes unequal population variances) Data Hypothesized Difference 0 Level of Significance 0.05 Population 1 Sample   Sample Size 30 Sample Mean XXXXXXXXXX Sample Standard Deviation 15.1154 Population 2 Sample   Sample Size 30 Sample Mean XXXXXXXXXX Sample Standard Deviation 17.2991 Intermediate Calculations Numerator of Degrees of Freedom XXXXXXXXXX Denominator of Degrees of Freedom 5.4313 Total Degrees of Freedom 56.9750 Degrees of Freedom 56 Standard Error 4.1942 Difference in Sample Means -1.0710 Separate-Variance t Test Statistic -0.2554 Two-Tail Test   Lower Critical Value -2.0032 Upper Critical Value 2.0032 p-Value 0.7994 Do not reject the null hypothesis   With (t=-0.25, p>0.05), I fail to reject the null hypothesis at 5% level of significance. There is no evidence to support the claim that the mean marks between two types of student status namely domestic and international differs significantly. The P value is equal to XXXXXXXXXXThere is 79.9% probability that the mean marks between two types of student status namely domestic and international differs significantly. Huck, S.W., Cormier, W.H. and Bounds, W.G., 1974.  Hypothesis 3 The null hypothesis, Ho: types of class and status of student is independent of each other. The alternative hypothesis, H1: types of class and status of student are dependent on each other. Lancaster, H.O., 1969.  Chi-square test of Independence is used to test the above hypothesis. The dependent variable is marks of the student. The independent variables are the type of class and this type of student status. For the application of Chi-square test of Independence, it is required that two variables are measured on the nominal scale of measurement. In this case, the type of student status and the type of class is a categorical variable measured by the nominal scale of measurement. The test statistic for chi-square test of Independence is given below. Lancaster, H.O., 1969.  Chi square=Sum {(Oi-Ei)^2/Ei} Where, Oi is the observed frequency Ei is the expected frequency Ei= (ri_total*ci_total)/(grand_total ) Degrees of
Answered Same Day Apr 04, 2020

Solution

Pooja answered on Apr 06 2020
145 Votes
Executive Summary
The objective of this report is to test if there is any significant difference in the marks of student on the basis of their gender, status and class. I also want to test if (gender, student status) as well as (gender, type of class) are independent of each other. Another area of concern is to test if mean marks of all students is greater than 50.  Lastly I want to know if the main effect of Gender and status as well as the interaction effect between them has a significant effect on marks. Techniques of hypothesis testing are used in order to analyze the data and make inferences.
Table of Contents
Executive Summary    1
Business Problem    3
Statistical Problem    4
Analysis    5
Different Classes    5
Output    5
Analysis    6
Different types of Students    8
Output    8
Analysis    9
Marks for different types of students    10
Output    10
Analysis    13
Hypothesis testing question 4    14
Output    14
Analysis    17
Hypothesis testing question 5    18
Output    18
Analysis    20
Hypothesis Question 6    21
Output    21
Analysis    22
Hypothesis Question 7    23
Output    23
Analysis    23
Conclusion    25
Business Problem
I want to analyze the marks of students on the basis of their gender, status and class.  The dependent variable is student marks.  The independent variables are gender, student status, and type of class. 
Statistical Problem
To test the difference of mean marks between Three Types of classes, the technique of descriptive statistics and one way anova is applied. To test the independence of Gender and student status, Chi square test of Independence is used. For testing the difference in mid marks for different type of students, T test for independent sample is explained. To know if the mean marks between domestic and international student status before I use T test for independent samples. One sample Z test is used to check if the mean marks of all students of greater than 50. To check the independence of gender and type of class, Chi square test of Independence is used. To check the significance of main effect of Gender and status as well as interaction effect between them, univariate analysis method is used.
Analysis
Different Classes
Output
    Descriptives
    MARK
    
    N
    Mean
    Std. Deviation
    Std. E
o
    95% Confidence Interval for Mean
    Minimum
    Maximum
    
    
    
    
    
    Lower Bound
    Upper Bound
    
    
    CLASS 1
    20
    70.8715
    15.39672
    3.44281
    63.6656
    78.0774
    50.84
    98.66
    CLASS 2
    20
    53.6605
    10.20395
    2.28167
    48.8849
    58.4361
    38.63
    73.30
    CLASS 3
    20
    43.8045
    8.37456
    1.87261
    39.8851
    47.7239
    26.18
    58.73
    Total
    60
    56.1122
    16.11478
    2.08041
    51.9493
    60.2751
    26.18
    98.66
    ANOVA
    MARK
    
    Sum of Squares
    df
    Mean Square
    F
    Sig.
    Between Groups
    7506.545
    2
    3753.272
    27.375
    .000
    Within Groups
    7814.942
    57
    137.104
    
    
    Total
    15321.487
    59
    
    
    
    Multiple Comparisons
    Dependent Variable: MARK
Tukey HSD
    (I) CLASS
    (J) CLASS
    Mean Difference (I-J)
    Std. E
o
    Sig.
    95% Confidence Interval
    
    
    
    
    
    Lower Bound
    Upper Bound
    CLASS 1
    CLASS 2
    17.21100*
    3.70276
    .000
    8.3006
    26.1214
    
    CLASS 3
    27.06700*
    3.70276
    .000
    18.1566
    35.9774
    CLASS 2
    CLASS 1
    -17.21100*
    3.70276
    .000
    -26.1214
    -8.3006
    
    CLASS 3
    9.85600*
    3.70276
    .027
    .9456
    18.7664
    CLASS 3
    CLASS 1
    -27.06700*
    3.70276
    .000
    -35.9774
    -18.1566
    
    CLASS 2
    -9.85600*
    3.70276
    .027
    -18.7664
    -.9456
    *. The mean difference is significant at the 0.05 level.
    MARK
    Tukey HSD
    CLASS
    N
    Subset for alpha = 0.05
    
    
    1
    2
    3
    CLASS 3
    20
    43.8045
    
    
    CLASS 2
    20
    
    53.6605
    
    CLASS 1
    20
    
    
    70.8715
    Sig.
    
    1.000
    1.000
    1.000
    Means for groups in homogeneous subsets are displayed.
    a. Uses Harmonic Mean Sample Size = 20.000.
Analysis
From the means lot of marks between Three Types of classes I observe that class 1 has the highest mean mark followed by class 2. Class 3 has the least mean marks.
From the table of descriptive statistics, the mean marks for class 1 is the highest with value of 70.8 along with the standard deviation of 15.39.  This is followed by the mean marks for class 2 with value of 53.6605 along with the standard deviation of 10.20395. Class 3 has the least mean marks with average of 43.8045 and standard deviation equal to 8.37456. The low value of standard deviation indicates that the mean is reliable.
I am 95% confident that estimated population mean marks for class 1 lies in an interval (63.6656, 78.0774).  I am 95% confident that estimated population mean marks for class 2 lies in an interval (48.8849, 58.4361).  I am 95% confident that estimated population mean marks for class 3 lies in the range of (39.8851, 47.7239).
Null hypothesis, ho:  there is significant difference in the mean marks between class 1, class 2 and class 3. The alternative hypothesis, H1: at least one of the mean marks between class 1 class 2 and class 3 differ significantly. With (F=27.37, p<0.05), I reject the null hypothesis 5% level of significance and conclude that at least one of the mean marks between class 1 class 2 and class 3 differ significantly.
The post Hoc analysis, Turkey Crammer test is used to know which pair of classes differ significantly with respect to the mean marks. There is a significant difference in the mean marks between (class1, class2), (class1, class3), and (class2, class3). 
Different types of Students
Output
    Case Processing Summary
    
    Cases
    
    Valid
    Missing
    Total
    
    N
    Percent
    N
    Percent
    N
    Percent
    GENDER * STATUS
    60
    100.0%
    0
    0.0%
    60
    100.0%
    GENDER * STATUS Crosstabulation
    
    STATUS
    Total
    
    DOMESTIC
    INTERNATIONAL
    
    GENDER
    FEMALE
    Count
    15
    15
    30
    
    
    Expected Count
    15.0
    15.0
    30.0
    
    MALE
    Count
    15
    15
    30
    
    
    Expected Count
    15.0
    15.0
    30.0
    Total
    Count
    30
    30
    60
    
    Expected Count
    30.0
    30.0
    60.0
    Chi-Square Tests
    
    Value
    df
    Asymp. Sig. (2-sided)
    Exact Sig. (2-sided)
    Exact Sig. (1-sided)
    Pearson Chi-Square
    .000a
    1
    1.000
    
    
    Continuity Co
ection
    .000
    1
    1.000
    
    
    Likelihood Ratio
    .000
    1
    1.000
    
    
    Fisher's Exact Test
    
    
    
    1.000
    .602
    Linear-by-Linear Association
    .000
    1
    1.000
    
    
    N of Valid Cases
    60
    
    
    
    
    a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 15.00.
    b. Computed only for a 2x2 table
Analysis
From the bar chart of Gender and status depicting their accounts, I...
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