Question

The sample data below have been collected based on a simple random sample from a normally distributed population. Complete parts a and b 3 7 4 Compute a 98% confidence interval estimate for the population mean. The 98% confidence interval for the population mean is from 1 to (Round to two decimal places as needed. Use ascending order.)

Answer 1

Solution:

The mean and standard deviation for the given data is:

$\dpi{100} \bar{x}=\frac{3+3+0+3+2+7+9+4+8+4}{10}=4.3$

$\dpi{100} s=\sqrt{\frac{\sum (x-\bar{x})^2}{n-1}}$

$\dpi{100} =\sqrt{\frac{(3-4.3)^2+(3-4.3)^2+(0-4.3)^2+...+(4-4.3)^2}{10-1}}$

$\dpi{100} =2.8304$

Now, the 98% confidence interval for the population mean is:

$\dpi{100} \bar{x} \pm t_{\frac{0.02}{2}}\left ( \frac{s}{\sqrt{n}} \right )$

Where:

$\dpi{100} t_{\frac{0.02}{2}}=2.821$ is the t critical value at 0.02 significance level for $\dpi{100} df=n-1=10-1=9$

Therefore, we have:

  $\dpi{100} 4.3\pm 2.821\left ( \frac{2.8304}{\sqrt{10}} \right )$

$\dpi{100} 4.3\pm 2.52$

$\dpi{100} \left ( 4.3-2.52,4.3+2.52 \right )$

$\dpi{100} \left (1.78,6.82 \right )$

Therefore, the 98% confidence interval for the population mean is $\dpi{100} \small \boldsymbol{\left (1.78,6.82 \right )}$