Question

Digital Viewing Times. According to marketing, the average time spent per day with digital media in $2010$was $3$hours and $14$minutes. For last year, a random sample of $20$American adults spent the following number of hours per day with digital media.

Find and interpret a $90\%$confidence interval for last year's mean time spent per day with digital media by American adults. (Note: $\overline{x}=5.16\mathrm{hr}$and $s=2.30\mathrm{hr}$)

Short answer

The $90\%$ confidence interval for American adults' mean daily time spent with digital media last year is $(4.271,6.049)$

Step-by-step solution

Given information

1.4	5.0	5.8	4.1	5.6
2.1	6.5	2.3	8.0	9.5
7.3	5.9	9.4	4.3	2.9
2.8	5.1	4.5	4.1	6.6

Concept

The formula used: The confidence interval$\overline{x}\pm t_{\frac{a}{2}}\left(\frac{s}{\sqrt{n}}\right)$

Calculation

Calculate the $90\%$confidence interval for American adults' average daily time spent with digital media last year.

Consider $\overline{x}=5.16,n=20$, and $s=2.30$

The needed value of $t_{\frac{a}{2}}$for $90\%$confidence with $19(=20-1)$degrees of freedom is $1.729$, according to "Table IV Values of $t_a$

Thus, the confidence interval is,

$$\begin{gathered} \overline{x}\pm t_{\frac{a}{2}}\left(\frac{s}{\sqrt{n}}\right)=5.16\pm 1.729\left(\frac{2.30}{\sqrt{20}}\right) \\ =5.16\pm 1.729(0.514) \\ =5.16\pm 0.889 \\ =(4.271,6.049) \end{gathered}$$

As a result, the $90\%$ confidence interval for American adults' mean daily time spent with digital media last year is $(4.271,6.049)$