Question

The burning rate of solid propellant used to power aircrew escape systems is normally distributed with standard deviation of 2 cm/s. A sample 64 items results in sample mean rate of 7 = 51.3 cm/s. Find the margin of error and 95% confidence interval for the average burning rate. Margin of Error: Confidence Interval: from to

Answer 1

The formula for calculating
(1 - a) 100%
confidence interval for true population mean $\mu$ is given by: $\bar{x}\pm Z_{\frac{\alpha}{2}}\left ( \frac{\sigma}{\sqrt{n}} \right )$

where,

Margin of error, E = Z2

Given: $\sigma=2\:cm/s;\:\bar{x}=51.3,\:cm/s;\:n=64$

Critical value: For 95% confidence,

a = 1-0.95 = 0.05

$Z_{\frac{\alpha}{2}}=Z_{\frac{0.05}{2}}=1.959964\approx \mathbf{1.96}$

$\mathrm{Margin\:of\:error,}\:E=Z_{\frac{\alpha}{2}}\left ( \frac{\sigma}{\sqrt{n}} \right )=1.96\left(\frac{2}{\sqrt{64}}\right)=\mathbf{0.49}$

Calculation of 95% confiednce interval for $\mu$ :

$\Rightarrow \bar{x}\pm Z_{\frac{\alpha}{2}}\left ( \frac{\sigma}{\sqrt{n}} \right )$

$\Rightarrow51.3\pm0.49$

$\Rightarrow(51.3-0.49,\:51.3+0.49)$

$\Rightarrow{\color{Blue} (50.81,\:51.79)}$

So the 95% confidence interval for the true population average $\mu$ is calculated as 50.81 to 51.79, i.e.,$\color{Blue} (50.81<\mu<51.79)$