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Q.8.8

Expert-verifiedFound in: Page 459

Book edition
OER 2018

Author(s)
Barbara Illowsky, Susan Dean

Pages
902 pages

ISBN
9781938168208

You do a study of hypnotherapy to determine how effective it is in increasing the number of hours of sleep subjects get each night. You measure hours of sleep for $12$ subjects with the following results. Construct a $95$% confidence interval for the mean number of hours slept for the population (assumed normal) from which you took the data. $8.2$; $9.1$; $7.7$; $8.6$; $6.9$; $11.2$; $10.1$; $9.9$; $8.9$; $9.2$; $7.5$; $10.5$

$\text{A}95\mathrm{\%}\mathrm{Cl}\text{on the population mean is}8.16\le \mu \le 9.80\text{.}$

Construct a $95$% confidence interval for the mean number of hours slept for the population (assumed normal) from which you took the data. $8.2$; $9.1$; $7.7$; $8.6$; $6.9$; $11.2$; $10.1$; $9.9$; $8.9$; $9.2$; $7.5$;

If $\stackrel{-}{x}$ and s are the mean and standard deviation of a random sample from a normal distribution with unknown variance ${\sigma}^{2}$,100(1-$\alpha $) % confidence interval on is given by

$\overline{x}-{t}_{\frac{\alpha}{2},n-1}\frac{s}{\sqrt{n}}\le \mu \le \overline{x}+{t}_{\frac{\alpha}{2},n-1}\frac{s}{\sqrt{n}}$ (1)

$\text{where}{t}_{\frac{\alpha}{2},n-1}\text{is the upper}100\frac{\alpha}{2}\text{percentage point of the}t\text{distribution with}n-1\text{degrees of freedom.}$

The simple mean is

$\overline{x}=\frac{1}{n}\sum _{i=1}^{n}{x}_{i}=\frac{8.2+9.1+\cdots +10.5}{12}=8.98$

and standard deviation is

$s={\left(\frac{\sum _{i=1}^{n}{({x}_{i}-\overline{x})}^{2}}{n-1}\right)}^{\frac{1}{2}}={\left(\frac{\sum _{i=1}^{12}{({x}_{i}-8.98)}^{2}}{11}\right)}^{\frac{1}{2}}=1.29$

$\text{Now, we need to find a}95\mathrm{\%}\mathrm{Cl}\text{on the population mean, then}$

$\frac{\alpha}{2}=\frac{1-0.95}{2}=0.025\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}{t}_{\frac{\alpha}{2},n-1}={t}_{0.025,11}=2.201.$ (2)

We used a probability table for the Student's t-distribution to find the value of t. The table gives t-scores that correspond to degrees of freedom (row) and the confidence level (column). The t-score is found where the row and column intersect in the table.

From Equation (1) and (2) we get

$8.98-2.2\frac{1.29}{\sqrt{12}}\le \mu \le 8.98+2.2\frac{1.29}{\sqrt{12}}$

Therefore,a $95$% cl on the population mean is

$8.16\le \mu \le 9.80$

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