Question

Table $8.2$shows a different random sampling of $20$cell phone models. Use this data to calculate a $93$% confidence interval for the true mean SAR for cellphones certified for use in the United States. As previously, assume that the population standard deviation is σ= $0.337$.

Phone Model	SAR	Phone Model	SAR
Blackberry pearl$8120$	$1.48$	Nokia E71x	$1.53$
HTC Evo Design 4G	$0.8$	Nokia N$75$	$0.68$
HTC Freestyle	$1.15$	Nokia N$79$	$1.4$
LG Ally	$1.36$	Sagem Puma	$1.24$
LG Fathom	$0.77$	Samsung Fascinate	$0.57$
LG Optimus Vu	$0.462$	Samsung Infuse 4G	$0.2$
Motorola Cliq XT	$1.36$	Samsung Nexus S	$0.51$
Motorola Droid pro	$1.39$	Samsung Replenish	$0.3$
Motorola Droid Razr M	$1.3$	Sony W5$18$a walkman	$0.73$
Nokia $7705$Twist	$0.7$	ZTE C$79$	$0.869$

Short answer

We estimate with $93$% confidence that the true population mean is between $0.7996$and $1.0724$

Step-by-step solution

Given Information

Given in the question, the value as

The population standard deviation is σ= 0.337.

Explanation

If $\stackrel{-}{x}$is the sample mean of a random sample of size n from a normal population with unknown variance $\sigma$^$2$, a $100$($1$-$\alpha$) % Cl on $\mu$is given by

$\overline{x}-z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}\le \mu \le \overline{x}+z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}$ (1)

$\text{where}{z}_{\frac{\alpha }{2}}\text{is the upper}100\frac{\alpha }{2}\text{percentage point of the standard normal distribution.}$

$\text{We know that standard deviation is}\sigma =0.337\text{, a random sample}n=20\text{and a sample mean}$

$\overline{x}=\frac{1.48+0.8+\cdots +0.869}{20}=0.936$

We need to find a $93$% confidence interval estimate for the population mean. Therefore,

$\frac{\alpha }{2}=\frac{1-0.93}{2}=0.035\Rightarrow z_{\frac{\alpha }{2}}=z_{0.035}=1.81$ (2)

Explanation

The earlier implication was obtained on a probability table for the standard normal distribution.

From equations (1) and (2) we get

$0.936-1.81\frac{0.337}{\sqrt{20}}\le \mu \le 0.936+1.81\frac{0.337}{\sqrt{20}}$

$\text{Therefore,}93\mathrm{\%}\mathrm{Cl}\text{for}\mu \text{is}$

$0.7996\le \mu \le 1.0724$