Question

given a sample mean value $overline{x}$, the critical $z$ values for the $98%$ confidence interval of the population mean $(mu)$ is $pm 2.33$. sixty - four people were asked how long they take in filling out a job application. the results showed a mean of $35$ minutes, with a standard deviation of $4.5$ minutes. what is the $98%$ confidence interval for the value of $mu$?
a. $33.69ltmult36.31$
b. $34.09ltmult35.91$
c. $34.53ltmult35.47$
d. $34.93ltmult35.07$

Explanation:

Step1: Calculate the margin of error

The formula for the margin of error \(E = z\times\frac{\sigma}{\sqrt{n}}\). Here, \(z = 2.33\), \(\sigma=4.5\), \(n = 64\) (since \(64\) people were asked). Then \(E=2.33\times\frac{4.5}{\sqrt{64}}=2.33\times\frac{4.5}{8}=2.33\times0.5625 = 1.31625\)

Step2: Calculate the confidence interval

The confidence interval is \(\bar{x}-E<\mu<\bar{x} + E\). Given \(\bar{x}=35\), so \(35 - 1.31625<\mu<35+1.31625\), which is \(33.68375<\mu<36.31625\)

Rounding to two decimal places, we get \(33.69<\mu<36.31\)

Answer:

A. \(33.69 <\mu<36.31\)