Suppose you wish to compare the means of six populations based on independent ra

Question

Suppose you wish to compare the means of six populations based on independent random samples, each of which contains 10 observations. The values of Total SS and SSE for the experiment are Total SS = 21.1 and SSE = 16.9. The sample means corresponding to populations 1 and 2 are x_1 = 3.05 and x_2^- = 2.46. (a) Find a 95% confidence interval for mu_1. (Round your answers to three decimal places.) to (b) Find a 95% confidence interval for the difference (mu_1 - mu_2). (Round your answers to three decimal places.) to

Explanation / Answer

SSr = SST - SSE

Given :

SST = 21.1

SSE = 16.9

SSR = 21.1-16.9 = 4.2

.

Sd = (SSR/n-1)

Sd = (4.2/9)

Sd = 0.6831

.

Answer to part a)

Now the formula of confidence interval of mean is :

x bar - t *Sd/n , xbar + t *Sd/n

Xbar = x1 = 3.05

t for 95% confidence interval with df = 9 is 2.262 ..............[We get this value from T table]

Sd = 0.6831

n = 10

.

on plugging these values in the formula we get

3.05 - 2.262 * 0.6831 /10 , 3.05 + 2.262 *0.6831 /10

2.561 and 3.539

.

Answer to part b)

s = (SSE/n1+n2-2)

s = (16.9/18)

s = 0.9690

.

The formula of confidence interval of difference of mean is :

(x1bar - x2bar) - t*s/(1/n1+1/n2) , (x1bar - x2bar) + t*s/(1/n1+1/n2)

on plugging the values we get:

(3.05-2.46) - 2.262 * 0.9690 * (1/10 +1/10) , (3.05-2.46) + 2.262 * 0.9690 * (1/10 +1/10)

0.59 - 2.262*0.969 *0.4472 , 0.59 + 2.262*0.9690 *0.4472

0.59 -0.9802 , 0.59 + 0.9802

-0.390 , 1.570

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