Answer :
Answer:
The 99 percent confidence interval for the proportion of all employees at the company who own an electric car is (0.0104, 0.0496).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
In a random sample of 500 employees, 15 reported owning an electric car.
This means that [tex]n = 500, \pi = \frac{15}{500} = 0.03[/tex]
99% confidence level
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.03 - 2.575\sqrt{\frac{0.03*0.97}{500}} = 0.0104[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.03 + 2.575\sqrt{\frac{0.03*0.97}{500}} = 0.0496[/tex]
The 99 percent confidence interval for the proportion of all employees at the company who own an electric car is (0.0104, 0.0496).
You can find the standard mean error and then use that to find the confidence interval limits.
The 99% confidence interval for the proportion of all employees at the company who own an electric car is given by
Lower limit = 0.0104
Upper limit = 0.0496
The confidence interval is (0.0104, 0.0496) (can include those end points too)
How to find the confidence interval for proportions?
Let the sample size be n
Let the interested quantities in the sample be present x times
Let the confidence level be of p%.
Then the level of significance is given by [tex]\alpha = 1 - p/100[/tex]
The standard mean error is calculated by:
[tex]SME = \sqrt{\dfrac{x(n-x)}{n^3}}[/tex]
Then the confidence interval limits can be found by
[tex]CI = \dfrac{x}{n} \pm Z_{\alpha/2}(SME)[/tex]
Using above formula for finding the confidence interval
We have n = 500, x = 15 and [tex]\alpha = 1 - 0.99 = 0.01[/tex]
Finding the standard mean error:
[tex]SME = \sqrt{\dfrac{x(n-x)}{n^3}}\\\\SME = \sqrt{\dfrac{15 \times 485}{500^3}} \approx 0.00763[/tex]
From the z tables, at [tex]\alpha/2 = 0.01/2 = 0.005[/tex], we get z = 2.5758
Thus, the confidence interval is obtained by
[tex]CI = \dfrac{x}{n} \pm Z_{\alpha/2}(SME)\\\\CI = \dfrac{15}{500} \pm 2.575 \times 0.00763\\\\CI = 0.03 \pm 0.0196[/tex]
Lower limit = 0.0104
Upper limit = 0.0496
Thus,
The 99% confidence interval for the proportion of all employees at the company who own an electric car is given by
Lower limit = 0.0104
Upper limit = 0.0496
The confidence interval is (0.0104, 0.0496) (can include those end points too)
Learn more about confidence interval here:
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