To calculate the margin of error (M.E.) for estimating a population mean, you can use the following formula:
\[ M.E. = t^* \times \frac{s}{\sqrt{n}} \]
where:
- \( t^* \) is the t-score corresponding to the desired confidence level and degrees of freedom,
- \( s \) is the sample standard deviation,
- \( n \) is the sample size.
Step 1: Identify the variables
- Sample size \( n = 19 \)
- Sample mean \( \bar{x} = 44.5 \) (not needed for M.E. calculation)
- Sample standard deviation \( s = 12.2 \)
- Confidence level = 98%
Step 2: Determine the degrees of freedom
\[ df = n - 1 = 19 - 1 = 18 \]
Step 3: Find the t-score
For a 98% confidence level and 18 degrees of freedom, we can use a t-table or calculator to find the critical value \( t^* \).
The t-score for \( 98% \) with \( 18 \) degrees of freedom is approximately \( 2.552 \).
Step 4: Calculate the standard error
\[ SE = \frac{s}{\sqrt{n}} = \frac{12.2}{\sqrt{19}} \approx \frac{12.2}{4.3589} \approx 2.8 \quad (\text{rounded to one decimal place}) \]
Step 5: Calculate the margin of error
\[ M.E. = t^* \times SE = 2.552 \times 2.8 \approx 7.142 \quad (\text{rounded to three decimal places}) \]
Step 6: Round to one decimal place
Rounding \( 7.142 \) to one decimal place, we get:
\[ M.E. \approx 7.1 \]
Thus, the margin of error \( M.E. \) is:
\[ \boxed{7.1} \]