Question

A nationwide survey of working adults indicates that only 50% of them are satisfied with their jobs. The president of a large company believes that more than 50% of employees at his company are satisfied with their jobs. To test his belief, he surveys a random sample of 100 employees, and 54 of them report that they are satisfied with their jobs. Use the = 0.05 level of significance.

Answer 1

Accept the null hypotheses( H₀:p=0.50) .This means only 50% of working adults are satisfied with their jobs.

Step-by-step explanation:

The hypotheses are;

H₀:p=0.50

Ha: p>0.50

where p is the working adults satisfied with their jobs

Check if normal model is a good fit for the sampling distribution as;

Check for condition using p=0.50

np=(100)(0.50) = 50

n(1-p) = 100(1-0.50)= 50

They are both more than 10 thus we can use the normal model to find p-value

Formula for z-score for a sample proportion is;

z=p'-p/√ p(1-p)/n

p' =x/n = 54/100 =0.54

z= 0.54-0.50 / √ 0.50(1-0.50)/100

z=0.04 / √ 0.0025

z=0.04 / 0.05 = 0.8

Use the z table to find the p-value where z=0.8

p-value, at z=0.8 = 0.788 but

find 1-p-value as 1-0.788 =0.212 ------because the alternative hypotheses is greater than thus we want the area to the right of z

p-value = 0.212

since p-value , 0.212 is more than 0.05 accept the null hypotheses.This means only 50% of working adults are satisfied with their jobs.