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 19 with a mean of 44.5 and a standard deviation of 12.2 at a confidence level of 98%.

Report ME accurate to one decimal place because the sample statistics are presented with this accuracy.
M.E. =

Answer 1

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:

\[
\boxed{7.1}
\]