Questions: If n=16, x̄=32, and s=4, construct a confidence interval at a 98% confidence level. Assume the data came from a normally distributed population. Give your answers to one decimal place. <μ<

If n=16, x̄=32, and s=4, construct a confidence interval at a 98% confidence level. Assume the data came from a normally distributed population.

Give your answers to one decimal place.

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Transcript text: If $\mathrm{n}=16, \bar{x}(\mathrm{x}$-bar $)=32$, and $s=4$, construct a confidence interval at a $98 \%$ confidence level. Assume the data came from a normally distributed population. Give your answers to one decimal place. $\square$ $<\mu<$ $\square$

Solution

Solution Steps

Step 1: Given Information

We are provided with the following parameters for constructing a confidence interval for the population mean:

Sample size: $ n = 16 $
Sample mean: $ \bar{x} = 32 $
Sample standard deviation: $ s = 4 $
Confidence level: $ 98\% $

Step 2: Determine the Significance Level

The significance level ($ \alpha $) is calculated as: \[ \alpha = 1 - \text{Confidence Level} = 1 - 0.98 = 0.02 \]

Step 3: Calculate the Margin of Error

To calculate the margin of error, we use the formula: \[ \text{Margin of Error} = t \cdot \frac{s}{\sqrt{n}} \] where $ t $ is the critical value from the t-distribution for $ n - 1 = 15 $ degrees of freedom at the $ 98\% $ confidence level. From the calculations, we find: \[ t \approx 2.6 \] Thus, the margin of error is: \[ \text{Margin of Error} = 2.6 \cdot \frac{4}{\sqrt{16}} = 2.6 \cdot \frac{4}{4} = 2.6 \]

Step 4: Construct the Confidence Interval

The confidence interval for the population mean is given by: \[ \bar{x} \pm \text{Margin of Error} \] Substituting the values, we have: \[ 32 \pm 2.6 \] This results in the interval: \[ (32 - 2.6, 32 + 2.6) = (29.4, 34.6) \]