To calculate the 98% confidence interval for the population mean, we'll use the t-distribution since the sample size is small (n < 30) and the population standard deviation is unknown.

Steps:

1. Calculate the sample mean ($\bar{x}$).
2. Calculate the sample standard deviation (s).
3. Find the t-critical value for a 98% confidence level and the degrees of freedom (df = n - 1).
4. Use the formula for the confidence interval: $\text{CI} = \bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$ Where:
   • $\bar{x}$ = sample mean
   • $t_{\alpha/2}$ = t-critical value
   • $s$ = sample standard deviation
   • $n$ = sample size

Step 1: Sample Mean ($\bar{x}$)

The sample data are: $11.8, 7.8, -2.5, 30.2, 5.7, 79.5, 5.6, 41.1$

$\bar{x} = \frac{11.8 + 7.8 + (-2.5) + 30.2 + 5.7 + 79.5 + 5.6 + 41.1}{8}$

$\bar{x} = \frac{178.2}{8} = 22.275$

So, the sample mean is approximately $22.28$.

Step 2: Sample Standard Deviation (s)

First, compute the squared differences from the mean for each sample value, then average them, and take the square root.

$s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}$

The squared differences:

$(11.8 - 22.275)^2 = 108.050625$ $(7.8 - 22.275)^2 = 207.515625$ $(-2.5 - 22.275)^2 = 614.05625$ $(30.2 - 22.275)^2 = 62.540625$ $(5.7 - 22.275)^2 = 277.485625$ $(79.5 - 22.275)^2 = 3231.45625$ $(5.6 - 22.275)^2 = 285.115625$ $(41.1 - 22.275)^2 = 348.55625$

Sum of squared differences: $108.05 + 207.52 + 614.06 + 62.54 + 277.49 + 3231.46 + 285.12 + 348.56 = 5115.80$

Now, compute the sample variance: $s^2 = \frac{5115.80}{7} = 731.83$

Take the square root to find the sample standard deviation: $s = \sqrt{731.83} \approx 27.04$

Step 3: t-Critical Value

For a 98% confidence interval with $n = 8$, the degrees of freedom (df) is: $df = n - 1 = 8 - 1 = 7$

From the t-distribution table, for a 98% confidence level and 7 degrees of freedom, the t-critical value ($t_{\alpha/2}$) is approximately 2.998.

Step 4: Calculate the Confidence Interval

Now, use the formula: $\text{CI} = \bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$ $\text{CI} = 22.28 \pm 2.998 \cdot \frac{27.04}{\sqrt{8}}$

Calculate the standard error: $\text{SE} = \frac{27.04}{\sqrt{8}} \approx \frac{27.04}{2.828} \approx 9.57$

Now calculate the margin of error: $\text{Margin of error} = 2.998 \cdot 9.57 \approx 28.69$

Thus, the confidence interval is: $\text{CI} = 22.28 \pm 28.69$

So the confidence interval is: $\text{CI} = (22.28 - 28.69, 22.28 + 28.69) = (-6.41, 50.97)$

Final Answer:

The 98% confidence interval for the population mean is: $\boxed{(-6.41, 50.97)}$

Let me know if you'd like more details or have any questions!

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Here are some related questions you might find interesting:

1. What is the difference between the t-distribution and the normal distribution?
2. How do you calculate the margin of error in a confidence interval?
3. What does it mean when we say a sample is normally distributed?
4. How does the sample size affect the width of a confidence interval?
5. How do you interpret the confidence level of 98%?

Tip: A smaller sample size increases the uncertainty in the estimate, which typically results in a wider confidence interval.