Temperature Confidence 7A1B89

1. **State the problem:** We have a sample of temperatures: $45.6, 51.8, 22.1, 41.3, 51.5, 15.8, 69.4, 31$. We want to find the 95% confidence interval for the population mean temperature assuming normal distribution. 2. **Formula for confidence interval:** $$\bar{x} \pm t_{\alpha/2, n-1} \times \frac{s}{\sqrt{n}}$$ where: - $\bar{x}$ is the sample mean - $t_{\alpha/2, n-1}$ is the t-score for 95% confidence and $n-1$ degrees of freedom - $s$ is the sample standard deviation - $n$ is the sample size 3. **Calculate sample size $n$:** $$n = 8$$ 4. **Calculate sample mean $\bar{x}$:** $$\bar{x} = \frac{45.6 + 51.8 + 22.1 + 41.3 + 51.5 + 15.8 + 69.4 + 31}{8} = \frac{328.5}{8} = 41.06$$ 5. **Calculate sample standard deviation $s$:** First, find squared deviations: $$\begin{aligned} (45.6 - 41.06)^2 &= 20.83 \\ (51.8 - 41.06)^2 &= 114.98 \\ (22.1 - 41.06)^2 &= 359.76 \\ (41.3 - 41.06)^2 &= 0.06 \\ (51.5 - 41.06)^2 &= 109.06 \\ (15.8 - 41.06)^2 &= 635.23 \\ (69.4 - 41.06)^2 &= 802.18 \\ (31 - 41.06)^2 &= 101.20 \end{aligned}$$ Sum of squared deviations: $$20.83 + 114.98 + 359.76 + 0.06 + 109.06 + 635.23 + 802.18 + 101.20 = 2143.30$$ Sample variance: $$s^2 = \frac{2143.30}{n-1} = \frac{2143.30}{7} = 306.19$$ Sample standard deviation: $$s = \sqrt{306.19} = 17.50$$ 6. **Find $t$-value for 95% confidence and 7 degrees of freedom:** $$t_{0.025,7} = 2.365$$ 7. **Calculate margin of error:** $$ME = t \times \frac{s}{\sqrt{n}} = 2.365 \times \frac{17.50}{\sqrt{8}} = 2.365 \times 6.19 = 14.64$$ 8. **Calculate confidence interval:** $$\begin{aligned} LB &= \bar{x} - ME = 41.06 - 14.64 = 26.42 \\ UP &= \bar{x} + ME = 41.06 + 14.64 = 55.70 \end{aligned}$$ 9. **Final answer:** The 95% confidence interval for the population mean temperature is: $$(26.42, 55.70)$$ The researcher is 95% confident that the population average temperature is within the temperature interval $(26.42, 55.70)$.