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 90% confidence level.

Answer 1

Answer:

The 90% confidence interval for the population mean is (300.78, 315.22).

Step-by-step explanation:

We have to calculate a 90% confidence interval for the mean.

The population standard deviation is know and is σ=17.

The sample mean is M=308.

The sample size is N=15.

As σ is known, the standard error of the mean (σM) is calculated as:

$\sigma_M=\dfrac{\sigma}{\sqrt{N}}=\dfrac{17}{\sqrt{15}}=\dfrac{17}{3.873}=4.389$

The z-value for a 90% confidence interval is z=1.645.

The margin of error (MOE) can be calculated as:

$MOE=z\cdot \sigma_M=1.645 \cdot 4.389=7.22$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 308-7.22=300.78\\\\UL=M+t \cdot s_M = 308+7.22=315.22$

The 90% confidence interval for the population mean is (300.78, 315.22).