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The mean salary of actuaries is LaTeX: \mu=\$100,000????=$100,000 and the standard deviation is LaTeX: \sigma=\$36,730????=$36,730. You are interested to know if the mean salary for women actuaries is different from the population mean, but are not sure if it is higher or lower. You set up a hypothesis test with level of significance LaTeX: \alpha=0.05????=0.05 and LaTeX: H_0:~~\mu=100,000H0: ????=100,000 LaTeX:H_a:~~\mu\neq100,000H????: ????≠100,000 You collect a simple random sample of LaTeX: n=36????=36 women's salaries and find that LaTeX: \bar{x}=\$111,000x¯=$111,000. Use this information for next three questions. What is the lower limit of this 95% confidence interval? (Round to the nearest whole number)

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Answer:

The lower limit of 95% confidence interval is 99002.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = $100,000

Sample mean, [tex]\bar{x}[/tex] = $111,000

Sample size, n = 36

Alpha, α = 0.05

Population standard deviation, σ = $36,730

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = \$100,000\\H_A: \mu \neq \$100,000[/tex]

We have to find the lower limit of the 95% confidence interval.

95% Confidence interval:

[tex]\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}[/tex]

Putting the values, we get,

[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]

[tex]111000 \pm 1.96(\dfrac{36730}{\sqrt{36}} )\\\\ = 111000 \pm 11998.4667 \\= (99001.5333,122998.4667)\\ \approx (99002,122999)[/tex]

The lower limit of 95% confidence interval is 99002.

The lower limit of this 95% confidence interval is  99002.

Given that,

  • a=0.05, Z(0.025) =1.96 (from stardard normal table)

Calculation:

The lower limit should be

[tex]\bar{x} - Z\timess\div vn \\\\= 111000-1.96\times36730\div \sqrt(36)[/tex]

= 99002

Find out more information about standard deviation here:

https://brainly.com/question/12402189?referrer=searchResults

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