Question

Assume that both populations are normally distributed. ​(a) Test whether mu 1 not equals mu 2μ1≠μ2 at the alpha equals 0.05α=0.05 level of significance for the given sample data.​(b) Construct a 9595​% confidence interval about mu 1 minus mu 2μ1−μ2.			Population 1	Population 2
		n	1717	1717
		x overbarx	10.810.8	14.214.2
		s	3.23.2	2.52.5

​(a) Test whether

mu 1 not equals mu 2μ1≠μ2

at the

alpha equals 0.05α=0.05

level of significance for the given sample data.

Determine the null and alternative hypothesis for this test.

A.

Upper H 0 :H0:mu 1 equals mu 2μ1=μ2

Upper H 1 :H1:mu 1 greater than mu 2μ1>μ2

B.

Upper H 0 :H0:mu 1 not equals mu 2μ1≠μ2

Upper H 1 :H1:mu 1 greater than mu 2μ1>μ2

C.

Upper H 0 :H0:mu 1 not equals mu 2μ1≠μ2

Upper H 1 :H1:mu 1 equals mu 2μ1=μ2

D.

Upper H 0 :H0:mu 1 equals mu 2μ1=μ2

Upper H 1 :H1:mu 1 not equals mu 2μ1≠μ2

Your answer is correct.

Detemine the​ P-value for this hypothesis test.

Pequals=nothing

​(Round to three decimal places as​ needed.)

Should it be rejected or not rejected?

Neeed Part B answered as well.

Answer 1

(A) Using TI 84 calculator

press stat then tests then 2-sampTTest

x1 = 10.8,s1 = 3.2 , n1 = 17

x2 = 14.2, s2 = 2.5 , n2 = 17

Pooled: No

press ENTER

p value = 0.002

p value is is less than 0.05 significance level, so we will reject Ho as there is significant difference between means

(B)

Using TI 84 calculator

press stat then tests then 2-sampTInt

x1 = 10.8,s1 = 3.2 , n1 = 17

x2 = 14.2, s2 = 2.5 , n2 = 17

c-level = 0.95

Pooled: No

press ENTER

(-5.411, -1.389)

confidence interval includes only negative values, so this also suggests to reject Ho