(SOLVED) hypothesis

Discipline: Finance

Academic Level: Undergrad. (yrs 3-4)

Paper Format: APA

Pages: 1 Words: 275

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Question

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 Gender F F F F F M M F M M F F M M M F F F M M M M M M F F M F M F F F M M F M F F F M F M M M B Age 18 17 19 19 20 20 20 17 19 19 19 19 19 19 19 20 20 20 17 20 19 19 20 20 19 20 19 19 20 20 19 19 19 17 20 20 19 20 19 19 17 20 19 20 C Time Spent in Studying (Hr) 4 9 10 14 4 7 12 22 1 17 25 15 3 1 22 40 25 26 22 5 10 10 16 7 7 15 20 10 22 25 14 1 1 22 7 22 10 12 25 22 22 16 20 7 D E F Exam Mark Under Financial Support? Attend any Tuition? 65 Yes No 58 No No 65 Yes No 73 No No 37 No No 44 No No 60 No No 88 Yes No 21 No No 84 No No 100 No No 75 No No 40 No No 40 No No 85 Yes No 80 No No 99 Yes No 90 Yes No 85 No No 32 No No 60 No No 65 No No 77 No Yes 48 No Yes 49 Yes Yes 70 No Yes 91 No Yes 21 No Yes 89 Yes Yes 80 No Yes 73 Yes Yes 21 No Yes 40 No Yes 85 No Yes 48 No Yes 78 Yes Yes 65 Yes Yes 60 No Yes 100 No Yes 95 No Yes 85 Yes Yes 77 No Yes 91 No Yes 48 Yes Yes 7 Section C (50 marks)
2 Conduct suitable hypothesis test(s) for each question. You may download the data file from
8 Polymal

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Expert Answer

Question1: To test whether there is an association between whether student has tution or not and whether the student has financial support or not.

First we need to compute contingency table for this

Financial support or not\Tuition or not	Yes	No	RowTotal
Yes	7	6	13
No	15	16	31
Column Total	22	22	44

The hypothesis are

Ho:Tution and financial support are independent

H1::Tution and financial support are not independent.

Steps:1 Compute expected values

Fromula=(Row total first*Column total first)/total

and similarly for otheres

Expected values
Financial support or not\Tuition	Yes	No
Yes	6.5	6.5
No	15.5	15.5

Step2: Compute Test-statistics

(Observed-Expected)^2/Expected
Financial support or not\Tuition	Yes	No
Yes	0.038462	0.038462
No	0.016129	0.016129

$2\nXe\n(One - Er,c)?\n=Σ\nEr,c\n= 0.10918\n$

The test statsitics follows chi-sqaure distribution with degrees of freedom=(r-1)*(c-1)=(2-1)*(2-1)=1

Chi-sqaure critival value is computed from chi-square table with degrees of freedom 1 and sigificance level 5%

Chi-critical=3.841

Conclusion: Here Chi-statistics value is less than chi-critical value, hence we fail to reject null hypothesis and conclude that Tution and financial support are independent

Question 2:

To test whether the average marks of students who has tution differs significantly from the average marks of students who do not have tuition.

x1	67.77273
x2	63
s1	22.99054
s2	23.08989
n1	22
n2	22

The hypothesis are

$Но : 1 = 2\n$ i.e the average marks of students who has tution equals to average marks of students who do not have tuition.

The test-statsitcs of the distribution is given by

$t=\nX1 – X2\n$12/nl + S22 /n2\n67.77 – 63\n22.992/22 +23.0892/22\n= 0.68703\n$

The test-statistics follows t-distribution with degrees of freedom

$(si?/n1 + S22 /n2)\nDF=\n= 41.99 = 42\n((S12 /n1)2)/(n1 – 1) + ((S22 /n2)2)/(n2 - 1)\n$

Since we have two tailed test P-value is computed from t-dsitribution table 5% level of significance with degrees of freedom

P-value=0.4958

Conclusion: Since P-value is greater than level of signiifcance, hence we fail to reject null hypothesis and conclude that the average marks of students who has tution equals to average marks of students who do not have tuition.

Question3:

a.) The correlation coefficent between time spent on study and "exam" mark is given by

$Correlation.coefficient\n(Xi – X) * (Yi-Y)\nVZ(Xi – X)2 * (Yi - Y)2\n0.7582\n$

which implies that there is strong linear relationship between time spent on study and "exam" marks.

b.) The hypothesis are

Ho:P=0 i.e Popuation correlation coefficent between time spent on study and "exam" marks is equal to zero

H1: P is not equal to zero

The test-statsitcs is

$r* n - 2\nt=\nV1-p2\n0.7282 * 44 – 2\n1 -0.75822\n7.5359\n$

The test-statistics follows t-distribution with segrees of freedom n-2=44-2=42

Since we have two tailed test P-value is computed from t-dsitribution table 5% level of significance with degrees of freedom

P-value=0.00001

Conclusion: Here P-value is less than significance level , hence we reject null hypothesis and conclude that there is statistically significance relationship between time spent on study and "exam" marks.

c.) The regression line for females to predict the "exam" marks given the time psent on study

"Exam" marks=a+b*Time spent on study+e

Using excel data analysis tool pack we have

Regression Statistics
Multiple R	0.615111
R Square	0.378362
Adjusted R Square	0.34728
Standard Error	19.09544
Observations	22

ANOVA
	df	SS	MS	F	Significance F
Regression	1	4438.737	4438.737	12.17307	0.002313
Residual	20	7292.717	364.6359
Total	21	11731.45

	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	41.94249	7.895524	5.312186	3.37E-05	25.47272	58.41227	25.47272	58.41227
Time Spent in Studying (Hr)	1.518319	0.435174	3.488992	0.002313	0.610562	2.426076	0.610562	2.426076

Hence the estimated regression line is

"Exam" marks=41.94+1.51*Time spent on study+e

d.) Since in queation, We have been asked to fit lieanr regression in sub part c only so we have only one regression line here as

"Exam" marks=41.94+1.51*Time spent on study+e

The predicted "exam" makrs of sally are

"Exam" marks=41.94+1.51*11=58.55