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...

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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                                Name: Tejas Kaja                                Student no: 5957989                                Campus: Wollongong       Executive Summary: With (F=27.37, p0.05), 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, p0.05), 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 0.7994. There 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

Pooja · Accepted Answer

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. Error
	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. Error
	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.

Data MARKGENDERCLASSSTATUS 84.84111 GENDER 1FEMALE 63.64121 2MALE 54.43131 55.65131 CLASS1CLASS 1 66.40121 2CLASS 2 74.51111 3CLASS 3 58.73131 43.64131...

Answer To: Data MARKGENDERCLASSSTATUS 84.84111 GENDER 1FEMALE 63.64121 2MALE 54.43131 55.65131...

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