1 Answer

Horgen · Answer 1 · 2022-01-03T14:37:59+0000

Complete Question

The complete question is shown on the first uploaded image

Answer:

The 99% confidence interval is
$2.4309 <  \mu <   4.9328$

Explanation:

From the question we are told that

The data is 3, 2,5, 3, 2, 6, 5,4.5, 3, 3, and 4

The sample size is n = 11

Generally the sample mean is mathematically represented as

$\= x  =  (\sum x_i)/(n)$

=>
$\= x  =  (3+ 2+5+ \cdots +4 )/(11)$

=>
$\= x  =  3.6818$

Generally the sample standard deviation is mathematically represented as

$\sigma  =  \sqrt{(\sum [ x_i - \= x ] )/(n) }$

=>
$\sigma  =  \sqrt{([ 3 - 3.6818 ]^2 +[ 2 - 3.6818 ]^2 + \cdots + [ 4 - 3.6818 ]^2   )/(11) }$

=>
$\sigma  =  1.3091$

Given that the confidence interval is 99% then the level of significance is mathematically represented as

$\alpha= (100 - 99) \%$

=>
$\alpha= 0.01$

Given that the sample size is small we will making use of t distribution table

Generally from the t distribution table the critical value of at a degree of freedom of is

$t_{(\alpha )/(2) , 10 } =3.16927267$

Generally the margin of error is mathematically represented as

E = 3.16927267 * (1.3091 )/(√(11) )

E = 1.251

Generally 99% confidence interval is mathematically represented as

3.6818 -1.251 < p < 3.6818 + 1.251

=>
$2.4309 <  \mu <   4.9328$

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