Questions: A sample of the length in inches for newborns is given below. Assume that lengths are normally distributed. Find the 99% confidence interval of the mean length. Length 15.7 21.5 19.2 16.2 16.5 15.3 18.6 21.7 16.7 22 Do not round in between steps. Round answers to at least 4 decimal places.

Solution Steps

The mean length \( \mu \) of the newborns is calculated using the formula:

\[ \mu = \frac{\sum_{i=1}^N x_i}{N} = \frac{183.4}{10} = 18.34 \]

Thus, the mean length is:

\[ \text{Mean length} = 18.34 \]

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

\[ \sigma^2 = \frac{\sum (x_i - \mu)^2}{n-1} = 6.9271 \]

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

\[ s = \sqrt{6.9271} = 2.6319 \]

Thus, the sample standard deviation is:

\[ \text{Sample standard deviation} = 2.6319 \]

To calculate the 99% confidence interval for the mean length, we use the formula:

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

Where:

• \( \bar{x} = 18.34 \)
• \( t \) (critical value for 99% confidence level with \( n-1 = 9 \) degrees of freedom) is approximately \( 3.2498 \)
• \( s = 2.6319 \)
• \( n = 10 \)

Calculating the margin of error:

\[ \text{Margin of Error} = 3.2498 \cdot \frac{2.6319}{\sqrt{10}} \approx 2.7048 \]

Thus, the confidence interval is:

\[ (18.34 - 2.7048, 18.34 + 2.7048) = (15.6352, 21.0448) \]

Final Answer