Question

Aaron sampled 101 students and calculated an average of 6.5 hours of sleep each night with a standard deviation of 2.14. Using a 96% confidence level, he also found that t* = 2.081.

confidence interval \(=\bar{x} \pm t^{\star} s / \sqrt{n}\)

A 96% confidence interval calculates that the average number of hours of sleep for working college students is between __________.

Answer 1

Given that

\(\mathrm{n}=101\)

\(\bar{x}=6.5\)

\(s=2.14\)

Note that, Population standard deviation \((\sigma)\) is unknown..So we use \(t\) distribution.

Our aim is to construct \(96 \%\) confidence interval.

\(\therefore c=0.96\)

\(\therefore \alpha=1-c=1-0.96=0.04\)

\(\therefore \alpha / 2=0.02\)

Also, d.f \(=\mathrm{n}-1=101-1=100\)

\(\therefore \mathrm{t}^{*}=t_{\alpha / 2, d . f .}=t_{\alpha / 2, n-1}=t_{0.02,100}=2.081\)

( use t table or \(\mathrm{t}\) calculator to find this value..)

The margin of error is given by

\(\mathrm{E}=\mathrm{t}^{*} *(s / \mathrm{n} \mathrm{n})\)

\(=2.081^{*}(2.14 / \sqrt{101})\)

\(=0.4431\)

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

\((\bar{x}-E)< \mu< (\bar{x}+E)\)

\((6.5-0.4431)< \mu< (6.5+0.4431)\)

\(6.0569<\mu< 6.9431\)

A \(96 \%\) confidence interval calculates that the average number of hours of sleep for working college students is between \(6.0569\) and \(6.9431\)