Question

A random sample of statistics students were asked to estimate the total number of hours they spend watching television in an average week. The responses are recorded in Table $8.4$. Use this sample data to construct a $98$% confidence interval for the mean number of hours statistics students will spend watching television in one week.

$0$	$3$	$1$	$20$	$9$
$5$	$10$	$1$	$10$	$4$
$14$	$2$	$4$	$4$	$5$

Table$8.4$

Short answer

A $98$% cl on the population mean is$2.396\le \mu \le 9.870$

Step-by-step solution

Step 1:Given Information

Use this sample data to construct a $98$% confidence interval for the mean number of hours statistics students will spend watching television in one week.

$0$	$3$	$1$	$20$	$9$
$5$	$10$	$1$	$10$	$4$
$14$	$2$	$4$	$4$	$5$

Explanation

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

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

The simple mean is

localid="1650530478779" $$\begin{gathered} \overline{x}=\frac{1}{n}\sum _{i=1}^n{x}_i \\ =\frac{0+3+1+\cdots +5}{15} \\ =6.13 \end{gathered}$$

and standard deviation is

localid="1650530496896" $$\begin{gathered} s={\left(\frac{\sum _{i=1}^n{(x_i-\overline{x})}^2}{n-1}\right)}^{\frac{1}{2}} \\ ={\left(\frac{\sum _{i=1}^{15}{(x_i-6.13)}^2}{14}\right)}^{\frac{1}{2}} \\ =5.51 \end{gathered}$$

Step 3: Explanation

Let's find 98% cl on the population mean

$$\begin{gathered} \frac{\alpha }{2}=\frac{1-0.98}{2} \\ =0.01 \\ \Rightarrow t_{\frac{\alpha }{2},n-1} \\ =t_{0.01,14} \\ =3.7368 \end{gathered}$$ (2)

To find the value of $t$, we used a probability table for the Student's $t-$distribution. The table shows $t-$scores for degrees of freedom (row) and confidence levels (column) (column). In the table, the$t-$score is obtained where the row and column intersect.

From Equations ($1$) and ($2$) we get

$6.13-3.7\frac{5.51}{\sqrt{15}}\le \mu \le 6.13+3.7\frac{5.51}{\sqrt{15}}$

$2.396\le \mu \le 9.870$