91Ó°ÊÓ

Let us consider,

Confidence intervals (Cls) of 90% and 99 percent for the percent additive in gasoline are required.

Solution

We'll use the following formula to figure out the confidence intervals: CI for

$\mathrm{μ}=\stackrel{→}{x}Â\pm \frac{ts}{\sqrt{n}}$

But first, let's figure out the necessary statistical parameters (mean and standard deviation) for calculating confidence intervals.

role="math" localid="1663302448239" $$\begin{gathered} samplemean\left(\stackrel{→}{x}\right)=\frac{i^{x_i}}{n} \\ =\frac{0.89}{6} \\ =0.1483 \end{gathered}$$

$$\begin{gathered} s=\sqrt{\frac{i^{{\left(x_i-\stackrel{→}{x}\right)}^2}}{n-1}} \\ s=\sqrt{\frac{{\left(0.13-0.1483\right)}^2+{(0.12-0.1483)}^2+{(0.16-0.1483)}^2+L+{\left(0.11-0.1483\right)}^2}{6-1}} \\ s=0.03430259465 \end{gathered}$$

Based on Table 4-4 in Chapter 4 of the book 4, t = 2.015 and 4.032 for the 90 % and 99 % levels of confidence, respectively. Substituting these values into equation (1) (formula of CI), together with the statistical values determined earlier, we get:

(a)$$\begin{gathered} 90\%CI=0.15Â\pm \frac{\left(2.105\right)\left(0.03430259465\right)}{\sqrt{6}} \\ =0.15Â\pm 0.03 \end{gathered}$$

(b)$$\begin{gathered} 99\%CI=0.15Â\pm \frac{\left(4.032\right)\left(0.03430259465\right)}{\sqrt{6}} \\ =0.15Â\pm 0.06 \end{gathered}$$

Note: To avoid erroneous final responses, values are only rounded off at the end of the calculation.

As a result, the percent additive in gasoline confidence ranges are $0.15Â\pm 0.03$and $0.15Â\pm 0.06$at the 90% and 99% levels of confidence, respectively.