1 Answer

Alican Balik · Answer 1 · 2021-07-11T10:15:05+0000

Answer:

The 99% confidence interval is
$165.776  <  \mu <  180.224$

Explanation:

From the question we are told that

The sample size is n = 100

The sample mean is
$\= x = 173 \ days$

The population standard deviation is
$\sigma = 28 \ days$

From the question we are told the confidence level is 99% , hence the level of significance is

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

=>
$\alpha = 0.01$

Generally from the normal distribution table the critical value of
$(\alpha )/(2)$ is

$Z_{(\alpha )/(2) } = 2.58$

Generally the margin of error is mathematically represented as

$E = Z_{(\alpha )/(2) } *  (\sigma )/(√(n) )$

E = 2.58 * (28 )/(√(100) )

=>
E = 7.224

Generally 95% confidence interval is mathematically represented as

$\= x -E <  \mu <  \=x  +E$

=>
$173 -7.224  <  \mu < 173 + 7.224$

=>
$165.776  <  \mu < 180.224$

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