Question

assume that a sample is used to estimate a population mean \\( \mu \\). find the margin of error m.e. that corresponds to a sample of size 21 with a mean of 21.6 and a standard deviation of 13.5 at a confidence level of 99%.
report me accurate to one decimal place because the sample statistics are presented with this accuracy.
\\( m.e. = \square \\)
answer should be obtained without any preliminary rounding. however, the critical value may be rounded to 3 decimal places.

Explanation:

Step1: Identify the distribution and formula

Since the population standard deviation is unknown and the sample size \( n = 21 \) (small sample, \( n<30 \)), we use the t - distribution. The formula for the margin of error (M.E.) when estimating a population mean with a t - distribution is \( M.E.=t_{\alpha/2,(n - 1)}\times\frac{s}{\sqrt{n}} \), where \( s \) is the sample standard deviation, \( n \) is the sample size, and \( t_{\alpha/2,(n - 1)} \) is the critical t - value.

Step2: Calculate the degrees of freedom and significance level

The degrees of freedom \( df=n - 1=21 - 1 = 20 \). The confidence level is \( 99\% \), so the significance level \( \alpha=1 - 0.99 = 0.01 \), and \( \alpha/2=0.005 \).

Step3: Find the critical t - value

We need to find \( t_{0.005,20} \). Looking up in the t - distribution table or using a calculator, \( t_{0.005,20}\approx2.845 \) (rounded to 3 decimal places as per the problem's instruction).

Step4: Substitute the values into the formula

We know that \( s = 13.5 \), \( n = 21 \), and \( t_{\alpha/2,(n - 1)}=2.845 \). First, calculate \( \frac{s}{\sqrt{n}}=\frac{13.5}{\sqrt{21}}\approx\frac{13.5}{4.5837}\approx2.945 \). Then, \( M.E.=2.845\times2.945\approx8.37 \). Rounding to one decimal place, we get \( M.E.\approx8.4 \).

Answer:

\( 8.4 \)