Question

Suppose germination periods, in days, for grass seed are normally distributed and have a known population standard deviation of 6 days and an unknown population mean. A random sample of 21 types of grass seed is taken and gives a sample mean of 34 days. Find the margin of error for the confidence interval for the population mean with a 99% confidence level.

Answer 1

Answer:3.725

Step-by-step explanation:

Formula of Margin of Error for (n<30):-

$E=t_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

Given : Sample size : n= 21

Level of confidence = 0.99

Significance level : $\alpha=1-0.99=0.01$

By using the t-distribution table ,

Critical value : $t_{n-1, \alpha/2}=t_{20,0.005}= 2.845$

Standard deviation: $\sigma=6$

Then, we have

$E=( 2.845)\dfrac{6}{\sqrt{21}}=3.724979386\approx3.725$

Hence, the margin of error for the confidence interval for the population mean with a 99% confidence level=3.725