In each of Exercises 10331038 we have provided summary stati

In each of Exercises 10.33–10.38, we have provided summary statistics for independent simple random samples from two populations. In each case, use the pooled t-test and the pooled t interval procedure to conduct the required hypothesis test and obtain the specied condence interval.

10.34) x¯1 = 10, s1 = 4, n1 = 15, x¯2 = 12, s2 = 5, n2 = 15

a. Two-tailed test, = 0.05

b. 95% condence interval

10.36) x¯1 = 20, s1 = 4, n1 = 10, x¯2 = 23, s2 = 5, n2 = 15

a. Left-tailed test, = 0.05

b. 90% condence interval

10.38) x¯1 = 20, s1 = 4, n1 = 30, x¯2 = 18, s2 = 5, n2 = 40

a. Right-tailed test, = 0.05

b. 90% condence interval

Solution

CI = x1 - x2 ± t a/2 * S * Sqrt ( 1 / n1 + 1 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=10
Standard deviation( sd1 )=4
Sample Size(n1)=15
Mean(x2)=12
Standard deviation( sd2 )=2
Sample Size(n1)=15
Value Pooled variance S^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
S^2 = (14*16 + 14*4) / (30- 2 )
S^2 = 10
S = 3.162
CI = [ ( 10-12) ± t a/2 * S * Sqrt( 1/15+1/15)]
= [ (-2) ± t a/2 * 3.162 * 0.3651 ]
= [ (-2) ± 2.145 * 1.1544 ]
= [-2 ± 2.4762] = [ -4.4762, 0.4762 ]