STATS Help! Calculate a 95% confidence interval for the true difference in means.

Two groups of rats [(a) normal and (b) adrenalectomized] were tested for blood viscosity. The
summary statistics are:
(a) (b)
sample size 11 9
sample mean 3.921 4.111
sample standard dev 0.527 0.601

(a) Calculate a 95% confidence interval for the true difference in means.
(b) Based on this confidence interval, would you conclude that the average blood viscosity differs
between normal and adrenalectomized rats? Explain your reasoning.
(c) Conduct a hypothesis test for a difference of means at the 5% level of significance. Is your
conclusion concordant with the conclusion in (b)?

_________________ (a) ______ (b)
sample size : _______ 11(n1) ___ 9(n2)
sample mean : _____ 3.921(x1bar) __ 4.111(x2bar)
sample standard dev : 0.527(s1) _ 0.601(s2)

a) (x1bar-x2bar) -/+ 2.1*Sqrt{[(n1-1)s1^2 + (n2-1)s2^2](n1+n2)/[(n1 + n2-2)n1n2]}
= (3.921-4.111) -/+ 2.1*Sqrt{[(11-1)0.527^2 + (9-1)0.601^2](11+9)/[(11 + 9-2)11*9]}
= -0.19 -/+ 0.5296
= (-0.7196, 0.3396)
The confidence interval for x2bar - x1bar is (-0.3396, 0.7196)
These confidence intervals contain '0'. Which means difference '0' is a possibility with 95% confidence. And so, we do not have enough evidence to reject Ho that there is no difference.
c) Test statistic t = -0.19/0.252 = -0.754. The critical values are -2.1 and 2.1. The value -0.754 is within the critical values and so, we do not have enough evidence to reject Ho that there is no difference.

jlynn -

Using a non-pooled two sample t-test and my trusty TI-84 calculator:

(a) CI = (-0.7317, 0.3517) with 16.13 degrees of freedom

(b) = It does not significantly differ because zero lies within the 95% confidence interval.

(c) you will get the same result ... do NOT reject the null hypothesis:

t = -0.743

p = 0.468 > 0.05, so do NOT reject.

Hope that helps