Questions: If X̄=79, S=14, and n=25, and assuming that the population is normally distributed, construct a 90% confidence interval estimate of the population mean, μ. ≤ μ ≤ (Round to two decimal places as needed.)

Solution Steps

We are provided with the following values:

• Sample mean $\bar{X} = 79$
• Sample standard deviation $S = 14$
• Sample size $n = 25$
• Confidence level = $90\%$

For a $90\%$ confidence level, the significance level $\alpha$ is calculated as: $$\alpha = 1 - 0.90 = 0.10$$ Since we are dealing with a small sample size and the population is assumed to be normally distributed, we will use the $t$-distribution. The critical value $t$ for $n - 1 = 24$ degrees of freedom at $\alpha/2 = 0.05$ is approximately $1.71$.

The standard error (SE) is calculated using the formula: $$SE = \frac{S}{\sqrt{n}} = \frac{14}{\sqrt{25}} = \frac{14}{5} = 2.8$$

The confidence interval is given by the formula: $$\bar{X} \pm t \cdot SE$$ Substituting the values: $$79 \pm 1.71 \cdot 2.8$$ Calculating the margin of error: $$1.71 \cdot 2.8 = 4.788$$ Thus, the confidence interval becomes: $$79 - 4.788 \leq \mu \leq 79 + 4.788$$ Calculating the bounds: $$74.212 \leq \mu \leq 83.788$$

Rounding to two decimal places, we have: $$(74.21, 83.79)$$

Final Answer