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Assignment - 1
Ans : 1
The null hypothesis for two sample mean is given by
The null hypothesis = ?0: ?1 − ?2 = Hypothesized Difference = 0
Alternative hypothesis = Ha : ?1 − ?2 ≠ 0
Or H a : μ 1 > μ 2

H a : μ 1 < μ 2
n1 = 15, n2= 18
α = o.o5
df = ?1 + ?2 − 2
= 15+18-2= 31
X1 bar =12.65
X2 bar =10.42

S.E = 0.75
t statistic for two sample mean is given by
t = [ (x1 - x2) - d ] / SE
t= 12.65-10.42/0.75 = 2.973
t Critical Value = 2.039
So from the above calculation we found that calculated value of t is greater than the critical value
So we reject the null hypothesis.
Ans : 2
The null hypothesis = ?0: ?1 = ?2
Alternative hypothesis = Ha : ?1 ≠ ?2
X1 bar = 30.45
X2 bar = 26.54
S.E = 2.15
n1 = 25, n2 = 27
df = 25+27-2 = 50
t = [ (x1 - x2) - d ] / SE
t = 30.45 – 26.54-0/2.15 = 1.818
  t critical = 2.009

As the t critical >t calculated
So from the above calculation we found that calculated value of t is less than the critical value
So, we accept the null hypothesis.

Ans 3.
The null hypothesis = ?0: ?1 = ?2
Alternative hypothesis = Ha : ?1 ≠ ?2
X1 bar = 4.12
X2 bar = 6.23
S.E = 1.44
n1 = 15 n2 = 13
df = 15+13-2 = 26
t = [ (x1 - x2) - d ] / SE
t = 4.12- 6.23/1.44 = -1.465
  t critical = 2.055

As the t critical >t calculated
So from the above calculation we found that calculated value of t is less than the critical value
So, the null hypothesis of equal population means accepted.

Ans 4:
Null hypothesis states that there is no difference between the two population means (i.e., d = 0), the null and alternative hypothesis are stated as follows.
Ho: μ1 = μ2
Ha: μ1 ≠ μ2
X1 bar = 7.34 s1 = 2.51 n1 = 14
X2 bar = 9.19 s2 = 3.78 n2 = 16
Df = 14+16-2= 28
Sp2 = (14-1) x (2.51)2 + (16-1) x (3.78)2 / 14+16-2
= 81.901 + 214.32/ 28
= 10.57
        
t = 7.34 – 9.19 -0/ sqrt 10.57(1/14 + 1/16)
= -1.85/1.415 = - 1.307
  t critical = 2.048
t critical value is larger than the absolute value of the calculated t of –1.307, so the...