Questions: Use the following sample to estimate a population mean μ. 48.2 61.2 64 48.9 18.9 Assuming the population is normally distributed, find the 98% confidence interval about the population mean. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places. 98% C.I. =

Solution Steps

The sample mean \( \mu \) is calculated using the formula:

\[ \mu = \frac{\sum_{i=1}^N x_i}{N} \]

For the given data:

\[ \mu = \frac{48.2 + 61.2 + 64 + 48.9 + 18.9}{5} = \frac{241.2}{5} = 48.24 \]

Thus, the sample mean is \( 48.24 \).

The sample variance \( \sigma^2 \) is calculated using the formula:

\[ \sigma^2 = \frac{\sum (x_i - \mu)^2}{n-1} \]

Calculating the variance gives:

\[ \sigma^2 = 319.4 \]

The sample standard deviation \( s \) is then:

\[ s = \sqrt{319.4} \approx 17.87 \]

Thus, the sample standard deviation is \( 17.87 \).

For a 98% confidence level, we use the formula for the confidence interval:

\[ \bar{x} \pm t \frac{s}{\sqrt{n}} \]

Where:

• \( \bar{x} = 48.24 \)
• \( t \) is the t-value for \( n-1 = 4 \) degrees of freedom at 98% confidence, which is approximately \( 3.75 \).
• \( s = 17.87 \)
• \( n = 5 \)

Calculating the margin of error:

\[ \text{Margin of Error} = t \frac{s}{\sqrt{n}} = 3.75 \cdot \frac{17.87}{\sqrt{5}} \approx 3.75 \cdot 7.99 \approx 29.96 \]

Thus, the confidence interval is:

\[ (48.24 - 29.96, 48.24 + 29.96) = (18.28, 78.20) \]

Final Answer