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Sagot :
The sample mean is 5.0833 and the standard deviation is 3.476 and the confidence interval is (2.87, 7.29).
In the given question, a random sample of 12 employees was taken, and the number of days each was absent for sickness was recorded (for a 1-year period).
Using these data to create a 95% confidence interval for the population mean days absent for sickness
2 5 3 7 10 0 6 8 5 11 3 1
a) We have to find the sample mean and the standard deviation (round to two decimal places).
The sample mean;
X = 2+5+3+7+10+0+6+8+5+11+3+1/12
X = 61/12
X = 5.0833
Now the standard deviation
SD = √[{x(1)-X)^2+(x(2)-X)^2…………….(x(12)-X)^2}/N]
After solving using that formula;
SD = 3.476
b) 95% confidence in interval for the population mean days absent for sickness.
Lower Bound = X - t(α/2)*s/√n
Upper Bound = X + t(α /2)*s/√n
where
α /2 = (1 - confidence level)/2 = 0.025
X = sample mean = 5.083333333
t(α/2) = critical t for the confidence interval = 2.20098516
s = sample standard deviation = 3.476108936
n = sample size = 12
df = n - 1 = 11
Thus,
Lower bound = 2.874719086
Upper bound = 7.291947581
Thus, the confidence interval is (2.87, 7.29).
To find the confidence interval link is here
brainly.com/question/22965427
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The right question is:
How healthy are the employees at direct marketing industry? A random sample of 12 employees was taken, and the number of days each was absent for sickness was recorded (for a 1-year period). Use these data to create a 95% confidence interval for the population mean days absent for sickness.
2 5 3 7 10 0 6 8 5 11 3 1
a) Find the sample mean and the standard deviation (round to two decimal places)
b) 95% confidence in interval for the population mean days absent for sickness.
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