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Part 1
Here are summary statistics for randomly selected weights of newborn​ girls: nequals36​, x overbarequals3180.6 ​g, sequals700.5 g. Use a confidence level of 95​% to complete parts​ (a) through​ (d) below.
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Part 1
a. Identify the critical value t Subscript alpha divided by 2 used for finding the margin of error.
t Subscript alpha divided by 2equals
  
2.03
​(Round to two decimal places as​ needed.)
Part 2
b. Find the margin of error.
Eequals
  
237.0 g
​(Round to one decimal place as​ needed.)
Part 3
c. Find the confidence interval estimate of mu.
  
enter your response here gless thanmuless than
  
enter your response here g
​(Round to one decimal place as​ needed.)

Let's go through each part step-by-step.

### Part (a) - Critical Value \( t_{\alpha/2} \)

You are given that the sample size is \( n = 36 \), and you want to calculate the critical value \( t_{\alpha/2} \) for a 95% confidence interval. Since you have already found this value as \( t_{\alpha/2} = 2.03 \), this is correct. Here's how you would normally find it:

1. Degrees of freedom \( df = n - 1 = 36 - 1 = 35 \).
2. Using a t-distribution table or calculator, find the critical value for \( \alpha/2 = 0.025 \) (since it's a two-tailed test for 95% confidence) with 35 degrees of freedom.
3. The result is \( t_{\alpha/2} = 2.03 \), which is rounded to two decimal places.

### Part (b) - Margin of Error (E)

The margin of error \( E \) is given by the formula:
\[
E = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}
\]
where:
- \( t_{\alpha/2} = 2.03 \) (from part a),
- \( s = 700.5 \) g (standard deviation),
- \( n = 36 \) (sample size).

Now, substitute these values into the formula:
\[
E = 2.03 \cdot \frac{700.5}{\sqrt{36}} = 2.03 \cdot \frac{700.5}{6} = 2.03 \cdot 116.75 = 236.9825
\]
Rounding to one decimal place:
\[
E \approx 237.0 \, \text{g}
\]

This matches the provided value.

### Part (c) - Confidence Interval Estimate of \( \mu \)

The confidence interval for the population mean \( \mu \) is given by:
\[
\text{CI} = \left( \bar{x} - E, \bar{x} + E \right)
\]
where:
- \( \bar{x} = 3180.6 \, \text{g} \) (sample mean),
- \( E = 237.0 \, \text{g} \) (margin of error from part b).

Now, calculate the lower and upper bounds of the confidence interval:
\[
\bar{x} - E = 3180.6 - 237.0 = 2943.6
\]
\[
\bar{x} + E = 3180.6 + 237.0 = 3417.6
\]
So, the 95% confidence interval is:
\[
2943.6 \, \text{g} < \mu < 3417.6 \, \text{g}
\]

### Final Answers:
- **Part (a):** \( t_{\alpha/2} = 2.03 \)
- **Part (b):** \( E = 237.0 \, \text{g} \)
- **Part (c):** \( 2943.6 \, \text{g} < \mu < 3417.6 \, \text{g} \)

---

Would you like more details on any step, or further explanation? 

Here are some related questions to expand your understanding:
1. How do degrees of freedom affect the critical value \( t_{\alpha/2} \)?
2. How does the sample size \( n \) influence the margin of error \( E \)?
3. What happens to the confidence interval if we change the confidence level to 99%?
4. Why do we use the \( t \)-distribution instead of the normal distribution in this case?
5. How would the margin of error change if the standard deviation was smaller?

**Tip:** The confidence interval becomes narrower as the sample size increases because the standard error decreases!

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.
Part (a): Identify the critical value t_{α/2}.
Part (b): Find the margin of error.
Part (c): Find the confidence interval estimate of μ.

Learn how to calculate a 95% confidence interval for the mean weight of newborn girls using t-distribution with n=36, x̄=3180.6g, and s=700.5g. This step-by-step guide covers finding the critical value, margin of error, and constructing the confidence interval.

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