Question

here are summary statistics for randomly selected weights of newborn girls: n = 36, \\( \overline { x } = 3211.1 \\) g, s = 689.4 g. use a confidence level of 90% 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 } = 1.69 \\)
(round to two decimal places as needed.)
b. find the margin of error.
e = 194.2 g
(round to one decimal place as needed.)
c. find the confidence interval estimate of \\( \mu \\)
\\( \square g < \mu < \square g \\)
(round to one decimal place as needed.)

Explanation:

Step1: Recall the formula for confidence interval

The confidence interval for the population mean \(\mu\) when the population standard deviation \(\sigma\) is unknown is given by \(\bar{x}-E <\mu<\bar{x} + E\), where \(\bar{x}\) is the sample mean and \(E\) is the margin of error.

Step2: Substitute the given values

We are given \(\bar{x}=3211.1\) g and \(E = 194.2\) g.

For the lower - bound:

$$ \bar{x}-E=3211.1- 194.2 $$

$$ =3016.9 $$

For the upper - bound:

$$ \bar{x}+E=3211.1 + 194.2 $$

$$ =3405.3 $$

Answer:

\(3016.9\) g\(<\mu<3405.3\) g