91Ó°ÊÓ

n=$20$, $\stackrel{¯}{x}=5.16\mathrm{hr}$and $s=2.30\mathrm{hr}$

The formula used: Lower confidence bound $=\stackrel{¯}{x}-t_{\mathrm{Î\pm }}\left(\frac{s}{\sqrt{n}}\right)$

Determine a $90\%$lower confidence bound for American adults' mean daily time spent with digital media last year.

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

From "Table IV Values of $t_a$" the required value of $t_{\frac{a}{2}}$for $90\%$confidence with $19(=20-1)$degrees of freedom is $1.328$

The lower confidence bound formula is as follows:

Lower confidence bound $=\stackrel{¯}{x}-t_{\mathrm{Î\pm }}\left(\frac{s}{\sqrt{n}}\right)$

$$\begin{gathered} \stackrel{¯}{x}-t_{\frac{a}{2}}\left(\frac{s}{\sqrt{n}}\right)=5.16-1.328\left(\frac{2.30}{\sqrt{20}}\right) \\ =5.16-1.328(0.5143) \\ =5.16-0.683 \\ =4.477 \end{gathered}$$

As a result, a $90\%$ lower confidence bound for last year's mean time spent per day by American adults with digital media is $4.477$ hours.

With $90\%$ confidence, the population mean time spent per day with digital median by American adults is larger than $4.477$ hours.

Part a: one-sided confidence interval:

Last year's mean time spent per day with digital media by American adults was $4.477$hours, according to the $90\%$lower confidence bound.

Exercise $8.130$'s two-sided confidence interval:

The $(4.271,6.049)$confidence interval for the mean time spent per day by American adults with digital media is $(4.271,6.049)$

The lower confidence bound for the mean time spent per day with digital median by American adults is clearly bigger than the lower confidence limit for the mean time spent per day with digital median by American adults, according to the findings.

That is,$Lowerbound(=4.477)>Lowerlimit(=4.271)$