IDNLearn.com: Your go-to resource for finding expert answers. Discover trustworthy solutions to your questions quickly and accurately with help from our dedicated community of experts.
Sagot :
90% confidence interval is 82.11 < μ < 93.89
"Information available from the question"
The complete question is this:
The managers of a company are worried about the morale of their employees. In order to determine whether a problem in this area exists, they decide to evaluate the attitudes of their employees with a standardized test. They select the Fortunato test of job satisfaction, which has a known standard deviation of 24 points.
Due to financial limitations, the managers decide to take a sample of 45 employees. This yields a mean score of 88 points. What is the 90% confidence interval?
Now, According to the question:
We have :
The confidence level = 0.90
The significance level, [tex]\alpha[/tex] = 0.10
The sample mean, x = 88
The population standard deviation, σ = 24
The sample size, n = 45
Critical value of z using the z - distribution table:
z(critical) = [tex]z_\alpha[/tex] /2
= [tex]z_0_._0_5[/tex]
= ±1.645
90% confidence interval:
μ = x ± z. σ/[tex]\sqrt{n}[/tex]
μ = 88 ± 1.645 x 24 / [tex]\sqrt{45}[/tex]
μ = 88 ± 5.89
82.11 < μ < 93.89
Hence, 90% confidence interval is 82.11 < μ < 93.89
Learn more about Confidence interval at:
https://brainly.com/question/20588426
#SPJ4
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Your questions deserve reliable answers. Thanks for visiting IDNLearn.com, and see you again soon for more helpful information.
