To calculate the 95% confidence interval for the mean using the t-score 𝑡

The formula to calculate the 95% confidence interval for the mean using the t-distribution is:

$CI = \left( \overline{x} - t \cdot \frac{s}{\sqrt{n}}, \, \overline{x} + t \cdot \frac{s}{\sqrt{n}} \right)$

Where:

• $\overline{x}$ = sample mean
• $t$ = t-score for the specified confidence level and degrees of freedom
• $s$ = sample standard deviation
• $n$ = sample size

Step-by-Step Calculation

1. Given Values:

   • Sample mean, $\overline{x} = 1.250$
   • Sample standard deviation, $s = 0.080$
   • Sample size, $n = 8$
   • t-score, $t = 2.365$

2. Calculate the Standard Error (SE): $SE = \frac{s}{\sqrt{n}} = \frac{0.080}{\sqrt{8}} = \frac{0.080}{2.828} \approx 0.0283$

3. Calculate the margin of error: $Margin\,of\,Error = t \cdot SE = 2.365 \cdot 0.0283 \approx 0.0669$

4. Calculate the Confidence Interval: $CI = \left( 1.250 - 0.0669, \, 1.250 + 0.0669 \right)$ $CI = \left( 1.1831, \, 1.3169 \right)$

Final Answer:

The 95% confidence interval for the mean is approximately $(1.1831, 1.3169)$.

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Would you like further clarification or details on any step? Here are some follow-up questions:

1. How is the t-score different from the z-score when calculating confidence intervals?
2. Why do we use $t$-distribution instead of $z$-distribution in this case?
3. How does increasing the sample size affect the width of the confidence interval?
4. Can you explain why the confidence interval may not always include the true population mean?
5. What happens to the margin of error if the sample standard deviation increases?

Tip: The t-distribution is used instead of the z-distribution when the sample size is small ($n < 30$) and/or the population standard deviation is unknown.