Question

23
multiple choice 4 points
construct the requested confidence interval.
an electronics company just finished designing a new tablet computer and is interested in estimating its battery - life. a random sample of 20 laptops with a full charge was tested and the battery - life was found to be approximately normal with a mean of 6 hours and a sample standard deviation of 1.5 hours. which of the following is the correct form for a 99% confidence interval?
$ci_{.99}(mu)=6pm2.576(0.3354)$
$ci_{.99}(mu)=6pm2.576(1.5)$
$ci_{.99}(mu)=6pm2.86(0.3354)$
$ci_{.99}(mu)=6pm2.86(1.5)$

Explanation:

Step1: Determine the critical value

Since the sample size \(n = 20\), the degrees of freedom \(df=n - 1=20 - 1 = 19\). For a \(99\%\) confidence interval, using the \(t\) - distribution table or a calculator, the critical value \(t_{\alpha/2}\) is \(t_{0.005,19}=2.86\) (because \(\alpha=1 - 0.99 = 0.01\) and \(\alpha/2=0.005\)).

Step2: Calculate the standard error

The formula for the standard error \(SE=\frac{s}{\sqrt{n}}\), where \(s = 1.5\) (sample standard deviation) and \(n = 20\). Then \(SE=\frac{1.5}{\sqrt{20}}\approx\frac{1.5}{4.472}\approx0.3354\).

Step3: Write the confidence interval formula

The formula for a confidence interval for the population mean \(\mu\) when the population standard deviation \(\sigma\) is unknown (we use the \(t\) - distribution) is \(CI=\bar{x}\pm t_{\alpha/2}\times SE\), where \(\bar{x} = 6\) (sample mean). Substituting the values, we get \(CI = 6\pm2.86\times0.3354\).

Answer:

\(CI_{.99}(\mu)=6\pm2.86(0.3354)\) (the third option)