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The principal randomly selected six students to take an aptitude test

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The principal randomly selected six students to take an aptitude test. Their scores were: 73.4 79.0 87.1 71.6 74.7 85.9 Determine a 90% confidence interval for the mean score for all students. a)73.10 < ? < 84.14 b)73.20 < ? < 84.04 c)84.04 < ? < 73.20 d)84.14 < ? < 73.10

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Solution:

Sample size n = 6

73.4, 79.0, 87.1, 71.6, 74.7, 85.9

Using calculator , first we find the sample mean $\bar x$ and sample standard deviation s

$\bar x$ = 78.616666666667

s = 6.5870833201552

Our aim is to construct 90% confidence interval.

$\therefore$ c = 0.90

$\therefore$ $\alpha$ = 1- c = 1- 0.90 = 0.10

$\therefore$ $\alpha$ /2 = 0.10 $\slash$ 2 = 0.05

Also, d.f = n - 1 = 5

$\therefore$ $t_{\alpha/2,d.f.}$ = $ta/2,1-1$ = $t$ _0.05,5 = 2.015

( use t table or t calculator to find this value..)

The margin of error is given by

E = $t$ $\alpha$ /2,d.f. * ( $s$ / $\sqrt{}$ n)

= 2.015 * (6.5870833201552 / $\sqrt{}$ 6 )

= 5.42

Now , confidence interval for mean( $\mu$ ) is given by:

( $\bar x$ - E ) < $\mu$ < ( $\bar x$ + E)

(78.616666666667 - 5.42) < $\mu$ < (78.616666666667 + 5.42)

73.20 < $\mu$ < 84.04

Option b) is correct.

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