1 Answer

Doolan · Answer 1 · 2021-10-30T16:27:57+0000

Answer:

The 90% confidence interval is
$87.95   <  \mu < 92.05$

Explanation:

From the question we are told that

The sample size is
n = 28

The sample mean is
$\= x = 90$

The standard deviation is
s= 60

Generally given that the sample size is not large enough i.e n < 28 then we will make use of the t distribution table

Generally the degree of freedom is mathematically represented as

df = n- 1

=>
df = 28 - 1

=>
df = 2 7

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=>
$\alpha = 0.10$

Generally from the student t distribution table the critical value of
$(\alpha )/(2)$ at a degree of freedom of
df = 2 7 is

$t_{(\alpha )/(2) , 27  } =  2.052$

Generally 95% confidence interval is mathematically represented as

$\= x -E <  \mu <  \=x  +E$

=>
$90  - 2.052  <  \mu < 90  + 2.052$

=>
$87.95   <  \mu < 92.05$

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