Compute two sample standard deviations

Consider the following data for two independent random samples taken from two normal populations.

Sample 1 14 26 20 16 14 18

Sample 2 18 16 8 12 16 14

a) Compute the two sample means and the two sample standard deviations.

b) What is the point estimate of the difference between the two population means?

c) Assuming α = .10, conduct p-value based and critical-value based hypothesis tests for the equality

of means of the two populations.

d) What is the 90% confidence interval estimate of the difference between the two population means?

How do the results compare in all the three approaches to hypothesis testing?

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Mean sample 1 = X1-bar = (14+26+20+16+14+18)/6 = 18

Mean sample 2 = X2-bar = (18+16+8+12+16+14)/6 = 14

Sample 1 SD = SD1

X1	X1-X1-bar	(X1-X1-bar)²
14	-4	16
26	8	64
20	2	4
16	-2	4
14	-4	16
18	0	0

Sum of (X1-X1-bar)² = 104

S1² = 104/6-1

= 20.8

SD1 = = 4.56

Sample 2 SD = SD2

X2	X1-X1-bar	(X1-X1-bar)²
18	4	16
16	2	4
8	-6	36
12	-2	4
16	2	4
14	0	0

Sum of (X2-X2-bar)² = 64

S2² = 64/6-1

= 12.8

SD2 = = 3.58

(b)

Point estimation of difference b/w two means = 18 - 14 = 4

(c)

t-test will be applied because sample size is small.

Hypothesis Formation

Null Hypothesis H₀: µ1 - µ2 = 0

Alternative Hypothesis H₁: µ1 - µ2 ≠ 0

t Statistic

t-statistic = (X1-bar - x2-bar)/S_p

Where S_P =

= 2.016

Critical value

Critical value of t with df=10 at 0.1 significance level = 1.812

Critical Region

Reject null hypothesis in favor of alternative if t is greater than t critical value of 1.812 or less than -1.812.

Computation

t-statistic = (18 - 14)/2.016

= 5.95

Decision

Null hypothesis is rejected in favor of alternative as Z value is greater than Z critical value.

(d)

90% CI of difference between means = (18-14) - 1.812*2.016

= 4 - 1.22 < µ < 4 + 1.22

= 2.78< µ< 5.22

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