Question

a. the sample mean, sample size, sample standard deviation, and confidence level given below. find a confidence interval for the mean of the population from which the sample was drawn
b. obtain the margin of error by taking half the length of the confidence interval
c. obtain the margin of error by using the formula ( t _ { alpha / 2 } cdot \frac { s } { sqrt { n } } )
( overline { x } = 24 )( n = 36 )( s = 4 )( ) confidence level ( = 98 % )
click here to view page 1 of the table of t-values with area alpha to its right
click here to view page 2 of the table of t-values with area alpha to its right

a. the ( 98 % ) confidence interval about ( mu ) is ( square ) to ( square )
(round to three decimal places as needed.)
b. the margin of error found by using the upper bound and the lower bound is, ( e = square )
(round to three decimal places as needed.)
c. the margin of error found by using above formula is, ( e = square )
(round to three decimal places as needed )

Explanation:

Step1: Find the critical value \( t_{\alpha/2} \)

For a \( 98\% \) confidence level, \( \alpha=1 - 0.98=0.02 \), and \( \alpha/2 = 0.01 \). The degrees of freedom \( df=n - 1=36 - 1 = 35 \). From the t - table, \( t_{0.01,35}\approx 2.438 \).

Step2: Calculate the margin of error using the formula \( E=t_{\alpha/2}\frac{s}{\sqrt{n}} \)

Substitute \( t_{\alpha/2}=2.438 \), \( s = 4 \), and \( n = 36 \) into the formula.
\( E=2.438\times\frac{4}{\sqrt{36}}=2.438\times\frac{4}{6}\approx1.625 \)

Step3: Calculate the confidence interval

The confidence interval is \( \bar{x}\pm E \). Given \( \bar{x}=24 \), the lower limit is \( 24 - 1.625=22.375 \) and the upper limit is \( 24+1.625 = 25.625 \)

Step4: Calculate the margin of error for part b

Since the confidence interval is \( (22.375,25.625) \), the length of the confidence interval is \( 25.625-22.375 = 3.25 \). The margin of error \( E=\frac{3.25}{2}=1.625 \)

Answer:

a. The \( 98\% \) confidence interval about \( \mu \) is \( 22.375 \) to \( 25.625 \)
b. The margin of error found by using the upper bound and the lower bound is \( E = 1.625 \)
c. The margin of error found by using the formula is \( E=1.625 \)