Question

A random sample of 45 restaurant servers found the average tips per shift were $150. Assume the sample standard deviation is $25. Using a 98% confidence level, calculate the following: Critical Value Group of answer choices

A. 2.414 B. 2.312 2 C.456 D.1.645

I cant seem use my z table correctly can anyone help.

Answer 1

Solution :

Given that,

sample size = n = 45

Degrees of freedom = df = n - 1 = 45 - 1 = 44

At 98% confidence level the t is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

t /2,df = t0.01,44 = 2.414

Critical Value = 2.414

option A. is correct