Question

Assume that a sample is used to estimate a population mean μ . Find the 98% confidence interval for a sample of size 73 with a mean of 29.4 and a standard deviation of 21.3. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places.98% C.I. = The answer should be obtained without any preliminary rounding.

Answer 1

Answer:
`98\text{ \% C.I. = \lparen23.601, 35.199\rparen}`

Step-by-step 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:

`\begin{gathered} \bar{x}\text{ }\pm\text{ Z}(s)/(√(n)) \\ where\text{ s= standard deviation} \\ \bar{x}\text{ = mean} \\ Z\text{ = 98\% z score} \end{gathered}`
`\begin{gathered} confidence\text{ interval = 29.4 }\pm\text{ Z }(21.3)/(√(73)) \\ Z\text{ = 98\% confidence = 2.326} \\ \\ Confidence\text{ interval = 29.4 }\pm\text{ 2.326 }*\text{ }(21.3)/(√(73)) \end{gathered}`
`\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}`
`98\text{ \% C.I. = \lparen23.601, 35.199\rparen}`