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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)/Sp

Where SP =

                  = 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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