Question

Suppose heights of seasonal pine saplings are normally distributed and have a known population standard deviation of 17 millimeters and an unknown population mean. A random sample of 15 saplings is taken and gives a sample mean of 308 millimeters. Find the confidence interval for the population mean with a 99%z0.10 z0.05 z0.025 z0.01 z0.0051.282 1.645 1.960 2.326 2.576

Answer 1

Answer:

$296.693\leq x\leq 319.307$

Step-by-step explanation:

The confidence interval for the population mean x can be calculated as:

$x'-z_{\alpha /2}\frac{s}{\sqrt{n} } \leq x\leq x'+z_{\alpha /2}\frac{s}{\sqrt{n} }$

Where x' is the sample mean, s is the population standard deviation, n is the sample size and $z_{\alpha /2}$ is the z-score that let a proportion of $\alpha /2$ on the right tail.

$\alpha$ is calculated as: 100%-99%=1%

So, $z_{\alpha/2}=z_{0.005}=2.576$

Finally, replacing the values of x' by 308, s by 17, n by 15 and $z_{\alpha /2}$ by 2.576, we get that the confidence interval is:

$308-2.576\frac{17}{\sqrt{15} } \leq x\leq 308+2.576\frac{17}{\sqrt{15} }\\308-11.307 \leq x\leq 308+11.307\\296.693\leq x\leq 319.307$