Math  /  Data & Statistics

Questionervals Quiz Question 7 of 13 This quiz: 13 point(s) possible This question: 1 point(s) possible Submit quiz
Here are summary statistics for randomly selected weights of newborn girls: n=36,x=3216.7 g, s=688.5 g\mathrm{n}=36, \overline{\mathrm{x}}=3216.7 \mathrm{~g}, \mathrm{~s}=688.5 \mathrm{~g}. Use a confidence level of 95%95 \% to complete parts (a) through (d) below. a. Identify the critical value tα/2\mathrm{t}_{\alpha / 2} used for finding the margin of error. tα/2=\mathrm{t}_{\alpha / 2}=\square (Round to two decimal places as needed.) b. Find the margin of error. E=gE=\square g (Round to one decimal place as needed.) c. Find the confidence interval estimate of μ\mu. \square g<μ<g<\mu< \square gg (Round to one decimal place as needed.) d. Write a brief statement that interprets the confidence interval. Choose the correct answer below.

Studdy Solution

STEP 1

What is this asking? We need to calculate a 95% confidence interval for the average weight of newborn girls given a sample of 36 girls with an average weight of 3216.7g and a standard deviation of 688.5g. Watch out! Don't forget to use the t-distribution since the population standard deviation is unknown, and we're dealing with a sample.
Also, pay close attention to rounding at each step!

STEP 2

1. Find the critical t-value.
2. Calculate the margin of error.
3. Construct the confidence interval.
4. Interpret the confidence interval.

STEP 3

Alright, first things first!
We're dealing with a **95%** confidence level, which means there's **5%** left over for the tails.
Since it's a two-tailed test, we split that 5% evenly, leaving **2.5%** in each tail.
So, we're looking for tα/2t_{\alpha/2}, where α/2=0.025\alpha/2 = 0.025.

STEP 4

We also know our sample size is **36**, so our degrees of freedom are n1=361=35n - 1 = 36 - 1 = \mathbf{35}.

STEP 5

Using a t-table or calculator with α/2=0.025\alpha/2 = \mathbf{0.025} and **35** degrees of freedom, we find our critical t-value: tα/22.03t_{\alpha/2} \approx \mathbf{2.03}.

STEP 6

The margin of error formula is E=tα/2snE = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}, where ss is the sample standard deviation and nn is the sample size.

STEP 7

Plugging in our values, we get E=2.03688.536=2.03688.562.03114.75232.8E = \mathbf{2.03} \cdot \frac{\mathbf{688.5}}{\sqrt{\mathbf{36}}} = 2.03 \cdot \frac{688.5}{6} \approx 2.03 \cdot 114.75 \approx \mathbf{232.8}.

STEP 8

The confidence interval is calculated as xˉ±E\bar{x} \pm E, where xˉ\bar{x} is the sample mean.

STEP 9

Our sample mean is 3216.7\mathbf{3216.7} and our margin of error is 232.8\mathbf{232.8}.

STEP 10

So, the confidence interval is 3216.7232.8<μ<3216.7+232.83216.7 - 232.8 < \mu < 3216.7 + 232.8, which simplifies to 2983.9<μ<3449.52983.9 < \mu < 3449.5.

STEP 11

We are 95% confident that the true average weight of newborn girls is between 2983.9g and 3449.5g.

STEP 12

tα/2=2.03t_{\alpha / 2} = 2.03 E=232.8gE = 232.8g 2983.9g<μ<3449.5g2983.9g < \mu < 3449.5gWe are 95% confident that the true average weight of newborn girls is between 2983.9g and 3449.5g.

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