Question

here are summary statistics for randomly selected weights of newborn girls: ( n = 36 ), ( \bar{x} = 3197.2 ) g, ( s = 692.6 ) 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 = square ) g
(round to one decimal place as needed.)

Explanation:

Step1: Recall the margin of error formula for t - distribution

The formula for the margin of error \( E \) when using the t - distribution is \( E=t_{\alpha/2}\times\frac{s}{\sqrt{n}} \), where \( t_{\alpha/2} \) is the critical value, \( s \) is the sample standard deviation, and \( n \) is the sample size.

Step2: Substitute the given values

We are given that \( t_{\alpha/2} = 1.69 \), \( s = 692.6 \) g, and \( n = 36 \). First, calculate \( \sqrt{n}=\sqrt{36}=6 \). Then, substitute the values into the formula: \( E = 1.69\times\frac{692.6}{6} \).

First, calculate \( \frac{692.6}{6}\approx115.4333 \). Then, multiply by \( 1.69 \): \( 1.69\times115.4333\approx195.1 \) (rounded to one decimal place).

Answer:

\( 195.1 \)