Math / Data & Statistics

QuestionHere are summary statistics for randomly selected weights of newborn girls: $n=36, \bar{x}=3216.7 \mathrm{~g}, \mathrm{~s}=688.5 \mathrm{~g}$ . Use a confidence level of $95 \%$ to complete parts (a) through (d) below. a. Identify the critical value $t_{\alpha / 2}$ 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. $E=$ $\square$ g (Round to one decimal place as needed.)

Studdy Solution

STEP 1

What is this asking? We need to find the margin of error for the average weight of newborn girls, given a sample, with 95% confidence. Watch out! Don't mix up the standard deviation of the sample with the standard deviation of the sampling distribution.
Also, remember to round correctly at each step!

STEP 2

1. Find the margin of error.

STEP 3

The margin of error $E$ is calculated using the formula: $E = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$ Where $t_{\alpha/2}$ is the critical t-value, $s$ is the sample standard deviation, and $n$ is the sample size.
This formula tells us how much the sample mean might differ from the true population mean.

STEP 4

We are given $t_{\alpha/2} = \textbf{2.03}$ , $s = \textbf{688.5}$ g, and $n = \textbf{36}$ .
Let's plug these values into our margin of error formula: $E = \textbf{2.03} \cdot \frac{\textbf{688.5}}{\sqrt{\textbf{36}}}$

STEP 5

First, let's simplify the square root: $\sqrt{36} = 6$ .
Now substitute that back into the formula: $E = \textbf{2.03} \cdot \frac{\textbf{688.5}}{6}$

STEP 6

Now, we divide $\textbf{688.5}$ by $6$ : $\frac{688.5}{6} = 114.75$ Finally, multiply this result by $\textbf{2.03}$ to find the margin of error: $E = \textbf{2.03} \cdot 114.75 = 232.7925$

STEP 7

The problem asks us to round to one decimal place.
So, our final margin of error is: $E \approx \textbf{232.8}$

STEP 8

The margin of error is $\textbf{232.8}$ g.

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Studdy Solution

STEP 1

STEP 2

1. Find the margin of error.

STEP 3

STEP 4

We are given $t_{\alpha/2} = \textbf{2.03}$ , $s = \textbf{688.5}$ g, and $n = \textbf{36}$ .
Let's plug these values into our margin of error formula: $E = \textbf{2.03} \cdot \frac{\textbf{688.5}}{\sqrt{\textbf{36}}}$

STEP 5

First, let's simplify the square root: $\sqrt{36} = 6$ .
Now substitute that back into the formula: $E = \textbf{2.03} \cdot \frac{\textbf{688.5}}{6}$

STEP 6

Now, we divide $\textbf{688.5}$ by $6$ : $\frac{688.5}{6} = 114.75$ Finally, multiply this result by $\textbf{2.03}$ to find the margin of error: $E = \textbf{2.03} \cdot 114.75 = 232.7925$

STEP 7

The problem asks us to round to one decimal place.
So, our final margin of error is: $E \approx \textbf{232.8}$

STEP 8

The margin of error is $\textbf{232.8}$ g.

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Studdy Solution

STEP 1

STEP 2

1. Find the margin of error.

STEP 3

STEP 4

We are given tα/2=2.03t_{\alpha/2} = \textbf{2.03}tα/2​=2.03, s=688.5s = \textbf{688.5}s=688.5 g, and n=36n = \textbf{36}n=36.Let's plug these values into our **margin of error** formula: E=2.03⋅688.536 E = \textbf{2.03} \cdot \frac{\textbf{688.5}}{\sqrt{\textbf{36}}} E=2.03⋅36​688.5​

STEP 5

First, let's simplify the square root: 36=6\sqrt{36} = 636​=6.Now substitute that back into the formula: E=2.03⋅688.56 E = \textbf{2.03} \cdot \frac{\textbf{688.5}}{6} E=2.03⋅6688.5​

STEP 6

Now, we divide 688.5\textbf{688.5}688.5 by 666: 688.56=114.75 \frac{688.5}{6} = 114.75 6688.5​=114.75 Finally, multiply this result by 2.03\textbf{2.03}2.03 to find the margin of error: E=2.03⋅114.75=232.7925 E = \textbf{2.03} \cdot 114.75 = 232.7925 E=2.03⋅114.75=232.7925

STEP 7

The problem asks us to round to one decimal place.So, our **final margin of error** is: E≈232.8 E \approx \textbf{232.8} E≈232.8

STEP 8

The margin of error is 232.8\textbf{232.8}232.8 g.

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We are given $t_{\alpha/2} = \textbf{2.03}$ , $s = \textbf{688.5}$ g, and $n = \textbf{36}$ .
Let's plug these values into our margin of error formula: $E = \textbf{2.03} \cdot \frac{\textbf{688.5}}{\sqrt{\textbf{36}}}$

First, let's simplify the square root: $\sqrt{36} = 6$ .
Now substitute that back into the formula: $E = \textbf{2.03} \cdot \frac{\textbf{688.5}}{6}$

Now, we divide $\textbf{688.5}$ by $6$ : $\frac{688.5}{6} = 114.75$ Finally, multiply this result by $\textbf{2.03}$ to find the margin of error: $E = \textbf{2.03} \cdot 114.75 = 232.7925$

The problem asks us to round to one decimal place.
So, our final margin of error is: $E \approx \textbf{232.8}$

The margin of error is $\textbf{232.8}$ g.