Questions: Here are summary statistics for randomly selected weights of newborn girls: n=36, x̄=3180.6 g, s=700.5 g. Use a confidence level of 95% to complete parts (a) through (d) below a. Identify the critical value tα/2 used for finding the margin of error. tα/2=2.03 (Round to two decimal places as needed.) b. Find the margin of error. E= g (Round to one decimal place as needed.)

Solution Steps

For a confidence level of \(95\%\), the critical value \( t_{\alpha/2} \) is given as \(2.03\). This value is used to calculate the margin of error for the sample mean.

The margin of error \( E \) is calculated using the formula:

\[ E = \frac{t_{\alpha/2} \times s}{\sqrt{n}} \]

where:

• \( t_{\alpha/2} = 2.03 \)
• \( s = 700.5 \, \text{g} \) (sample standard deviation)
• \( n = 36 \) (sample size)

Substituting the values, we have:

\[ E = \frac{2.03 \times 700.5}{\sqrt{36}} \]

\[ E = \frac{2.03 \times 700.5}{6} \]

\[ E = \frac{1421.015}{6} \]

\[ E = 236.8358 \]

Rounding to one decimal place, the margin of error is \( 236.8 \, \text{g} \).

Final Answer