Question

For a group of 10 men subjected to a stress situation, the mean number of heartbeats per minute was 126 and the standard deviation was 4. Find the 95% confidence interval of the true mean. Populations seems to be normally distributed.

Please round to the nearest whole number and write your answer in the form ( , ); for example (31,34) or (25,46). Do not type in any extra spaces

Answer 1

To calculate the 95% confidence interval for the true mean, we can use the formula for the confidence interval:

\[
\text{Confidence Interval} = \bar{x} \pm t \times \left( \frac{s}{\sqrt{n}} \right)
\]

where:
- \(\bar{x}\) is the sample mean
- \(t\) is the t-value for the desired confidence level (from the t-distribution table)
- \(s\) is the sample standard deviation
- \(n\) is the sample size

Given that:
- \(\bar{x} = 126\)
- \(s = 4\)
- \(n = 10\)

**Step 1:** Find the t-value for a 95% confidence interval with \(n - 1 = 10 - 1 = 9\) degrees of freedom. For 95% confidence and 9 degrees of freedom, the t-value is approximately 2.262 (this value can be found in a t-distribution table).

**Step 2:** Calculate the standard error (SE):

\[
SE = \frac{s}{\sqrt{n}} = \frac{4}{\sqrt{10}} \approx \frac{4}{3.1623} \approx 1.2649
\]

**Step 3:** Calculate the margin of error (ME):

\[
ME = t \times SE \approx 2.262 \times 1.2649 \approx 2.8614
\]

**Step 4:** Calculate the confidence interval:

\[
\text{Confidence Interval} = 126 \pm 2.8614
\]
\[
= (126 - 2.8614, 126 + 2.8614) \approx (123.1386, 128.8614)
\]

Rounding to the nearest whole number gives us:

\[
(123, 129)
\]

Thus, the 95% confidence interval for the true mean is:

(123,129)