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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
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