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.)

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.)

Transcript text: If $\bar{X}=79, \mathrm{~S}=14$, and $\mathrm{n}=25$, and assuming that the population is normally distributed, construct a $90 \%$ confidence interval estimate of the population mean, $\mu$. $\square$ $\leq \mu \leq$ $\square$ (Round to two decimal places as needed.)

Solution

Solution Steps

Step 1: Given Information

We are provided with the following values:

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

Step 2: Determine the Critical Value

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 $.

Step 3: Calculate the Standard Error

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

Step 4: Construct the Confidence Interval

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: \[ 74.212 \leq \mu \leq 83.788 \]

Step 5: Round the Results

Rounding to two decimal places, we have: \[ (74.21, 83.79) \]