Assume that a sample is used to estimate a population mean . Find the margin of error M.E.
that corresponds to a sample of size 7 with a mean of 81.5 and a standard deviation of 13.6 at
a confidence level of .
Report ME accurate to one decimal place because the sample statistics are presented with
this accuracy.
M.E.
Answer should be obtained without any preliminary rounding. However, the critical value may
be rounded to 3 decimal places.

Ask by Mccoy Delgado. in the United States

Mar 29,2025

Question

Assume that a sample is used to estimate a population mean . Find the margin of error M.E.
that corresponds to a sample of size 7 with a mean of 81.5 and a standard deviation of 13.6 at
a confidence level of .
Report ME accurate to one decimal place because the sample statistics are presented with
this accuracy.
M.E.
Answer should be obtained without any preliminary rounding. However, the critical value may
be rounded to 3 decimal places.

Ask by Mccoy Delgado. in the United States

Mar 29,2025

UpStudy AI · Accepted Answer

We use the formula
$$
\text{M.E.} = t_{\frac{\alpha}{2},\,n-1}\frac{s}{\sqrt{n}},
$$
where $$ t_{\frac{\alpha}{2},\,n-1} $$ is the critical value from the $$ t $$-distribution with $$ n-1 $$ degrees of freedom.

1. The sample size is $$ n=7 $$, so the degrees of freedom are
   $$
   n - 1 = 6.
   $$

2. The confidence level is $$ 99.9\% $$; therefore,
   $$
   \alpha = 1 - 0.999 = 0.001 \quad \text{and} \quad \frac{\alpha}{2} = 0.0005.
   $$
   From the $$ t $$-table (rounding the critical value to 3 decimal places), we have
   $$
   t_{0.0005, 6} \approx 4.03.
   $$

3. The standard error (SE) is
   $$
   \text{SE}=\frac{s}{\sqrt{n}} = \frac{13.6}{\sqrt{7}}.
   $$
   Compute the square root:
   $$
   \sqrt{7} \approx 2.6458,
   $$
   so
   $$
   \frac{13.6}{2.6458} \approx 5.1396.
   $$

4. Now, the margin of error is
   $$
   \text{M.E.} = 4.03 \times 5.1396 \approx 20.717.
   $$
   Rounding to one decimal place (without any preliminary rounding in the calculation steps) yields
   $$
   \text{M.E.} \approx 20.7.
   $$

Thus, the margin of error is $$ \boxed{20.7} $$.
The margin of error is 20.7.

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