Can you please show the work? 2. You want to test the following hypotheses using
ID: 3046816 • Letter: C
Question
Can you please show the work?
2. You want to test the following hypotheses using a level of significance of 5%:
H0: 2 1 = 30.0 H1: 2 1 30.0
The sample data are:
n1 = 9, x1 = 204.3, S 2 1 = 1,242
n2 = 15, x2 = 213.7, S 2 2 = 1,605.
Each sample comes from a population with a normal distribution, but there is no reason to
believe that their population variances are equal.
a. What is the estimated variance of x2 x1 ?
b. What is the estimated standard error of x2 x1.
c. What is the attained value of the test statistic?
d. How many degrees of freedom should you use with this test statistic?
e. What is the p value?
f. Can we reject the null hypothesis?
Explanation / Answer
a) Estimated Variance of X2bar-X1bar is
Sp= the common standard deviation can be estimated by the pooled standard deviation
= Sp= sqrt(((n1-1)*s1^2+(n2-1)*s2^2)/(n1+n2-2))
= Sqrt((9-1)*1242+(15-1)*1605/9+15-2)
=Sqrt((8*1242+14*1605/22)
=Sqrt(9936+22470/22)
=Sqrt(32406/22)
=Sqrt(1473)= 38.38
Estimated Variance = 1473
b) estimates standard error when variances are equal= S.E(x1bar-x2bar)= sqrt(sp^2(1/n1+1/n2))
= Sqrt (1473*(1/9+1/15))
=sqrt(1473*(0.11+0.067)
=sqrt(1473*0.177)
=Sqrt (260.721)
=16.15
Estimated error when variances are Not equal
S.E= sqrt(s1^2/n1+S2^2/n2)
= Sqrt (1242/9+1605/15)
= Sqrt(138+107)
=Sqrt(245)
= 15.65
c) Since population Variance are not equal
t Statistic = x1bar-x2bar/S.E
= 204.3-213.7-30/15.65
=-39.4 /15.65
= -2.52
The P-Value is .019495.The result is significant at p < .05.
Degree of freedom= n1+n2-2= 9+15-2= 22
Yes we can reject null hypothesis H0