Question

Use the information below to determine whether.  State the statistical hypotheses and report the P-value.  Find the 95% confidence intervals.

Sample 1: N1=48, X1= 4.99, ss1= 13.2

Sample 2: N2= 32, X2= 5.83, ss2= 29.2

Answer 1

Hypothesis testing:

The null hypothesis, ho: there is no significant difference in the mean of sample 1 and 2. u1 =/= u2

An alternative hypothesis, ho: there is a significant difference in the mean of sample 1 and 2. u1=u2

	SAMPLE 1	SAMPLE 2
n=	48	32
mean=	4.99	5.83
s=	0.5300	0.9705
s^2/n	0.0059	0.0294

test statistic, t =    = (Xbar1 - Xbar2) / sqrt(s1^2/n1 + s2^2/n2)
t =    = (4.99 - 5.83) / sqrt(0.0059+0.0294)
t =    -4.4709

df=   (s1^2/n1 + s2^2/n2)^2 / ((s1^2/n1)^2/(n1-1) + (s2^2/n2)^2/(n2-1))
df=   (0.0059+0.0294)^2 / ( 0.0059^2/(48-1) + 0.0294^2/(32-1) )
df=   44

p-value =    2*(1-P(T<|t|)
p-value =    2*(1-P(T<abs(-4.4709))
p-value =    T.DIST.2T(abs(-4.4709),44)
p-value =    5.4105E-05 = 0.00005

With t=-4.47, p<5% i reject ho and conclude that there is a significant difference in the mean of sample 1 and 2. u1=u2

95% confidence interval:
t(a/2, df)   =t(0.05/2,44)
t(a/2, df)   =t.INV.2T(0.05,44) =
t(a/2, df)   2.0154

CI = (xbar1 - xbar2) +- t(a/2,dF)*Sqrt(s1^2/n1 + s2^2/n2)
lower = (4.99-5.83)-2.0154*SQRT(0.0059+0.0294) = -1.2187
upper = (4.99-5.83)+2.0154*SQRT(0.0059+0.0294) = -0.4613