Questions: Here are summary statistics for randomly selected weights of newborn girls: n=36, x̄=3227.8 g, s=686.4 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=233 g (Round to one decimal place as needed.) c. Find the confidence interval estimate of μ g < μ < g 79 (Round to one decimal place as needed.)

Here are summary statistics for randomly selected weights of newborn girls: n=36, x̄=3227.8 g, s=686.4 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=233 g (Round to one decimal place as needed.) c. Find the confidence interval estimate of μ g < μ < g 79 (Round to one decimal place as needed.)
Transcript text: Here are summary statistics for randomly selected weights of newborn girls: $\mathrm{n}=36, \bar{x}=3227.8 \mathrm{~g}, \mathrm{~s}=686.4 \mathrm{~g}$. Use a confidence level of $95 \%$ to complete parts (a) through ( d ) below. a. Identify the critical value $\mathrm{t}_{\alpha / 2}$ used for finding the margin of error. \[ t_{\alpha / 2}=2.03 \] (Round to two decimal places as needed) b. Find the margin of error. \[ E=233 \mathrm{~g} \] (Round to one decimal place as needed.) c. Find the confidence interval estimate of $\mu$ $\square$ $\mathrm{g}<\mu<\square$ $\square$ 79 (Round to one decimal place as needed.)
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Solution

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Solution Steps

Step 1: Identify the Critical Value

The critical value \( t_{\alpha/2} \) for a 95% confidence level is given as:

\[ t_{\alpha/2} = 2.03 \]

Step 2: Calculate the Margin of Error

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

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

Substituting the values:

\[ E = 2.0 \cdot \frac{686.4}{\sqrt{36}} = 224.2 \, \text{g} \]

Step 3: Find the Confidence Interval Estimate of \( \mu \)

The confidence interval for the mean \( \mu \) is given by:

\[ \bar{x} \pm E \]

Calculating the lower and upper bounds:

\[ \text{Lower Bound} = 3227.8 - 224.2 = 3003.6 \, \text{g} \] \[ \text{Upper Bound} = 3227.8 + 224.2 = 3452.0 \, \text{g} \]

Thus, the confidence interval is:

\[ 3003.6 \, \text{g} < \mu < 3452.0 \, \text{g} \]

Final Answer

The answers to the sub-questions are:

  • Critical value \( t_{\alpha/2} = 2.03 \)
  • Margin of error \( E = 224.2 \, \text{g} \)
  • Confidence interval \( 3003.6 \, \text{g} < \mu < 3452.0 \, \text{g} \)

\[ \boxed{t_{\alpha/2} = 2.03, \, E = 224.2 \, \text{g}, \, 3003.6 \, \text{g} < \mu < 3452.0 \, \text{g}} \]

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