Respuesta :
Answer: [tex]98\text{ \% C.I. = \lparen23.601, 35.199\rparen}[/tex]Explanation:
Given:
sample size = 73
mean = 29.4
standard deviation = 21.3
To find:
98% confidence interval for the sample size and mean
To get the confidence interval, we'll apply the formula:
[tex]\begin{gathered} \bar{x}\text{ }\pm\text{ Z}\frac{s}{\sqrt{n}} \\ where\text{ s= standard deviation} \\ \bar{x}\text{ = mean} \\ Z\text{ = 98\% z score} \end{gathered}[/tex][tex]\begin{gathered} confidence\text{ interval = 29.4 }\pm\text{ Z }\frac{21.3}{\sqrt{73}} \\ Z\text{ = 98\% confidence = 2.326} \\ \\ Confidence\text{ interval = 29.4 }\pm\text{ 2.326 }\times\text{ }\frac{21.3}{\sqrt{73}} \end{gathered}[/tex][tex]\begin{gathered} Confidence\text{ interval = 29.4 }\pm\text{ 5.7987} \\ \\ Conf\imaginaryI dence\text{ }\imaginaryI\text{nterval = 29.4 +5.7987 or 29.4 - 5.7987} \\ \\ Confidence\text{ interval = 35.1987 or 23.6013} \\ \\ To\text{ 3 decimal place, upper bound = 35.199 and lower bound = 23.601} \end{gathered}[/tex][tex]98\text{ \% C.I. = \lparen23.601, 35.199\rparen}[/tex]