Question

1. The following is a simple random sample taken from a population of employees in a company. The information that was asked of each person was the monthly salary they had last year measured in dollars. 1200 4500 1340 3100 2500 5100 2100 1300 2300 4100 3200 1200 5200 3500 2600 (a) Determine the sample average wage and the sample standard deviation of the wages of the employees. (b) Determine the standard error of the mean assuming the population is infinite. Find the 95% confidence interval for the average salary of employees. (c) Assuming that the company has a total of 50 employees. How should I change the interval reliable for the average salary of employees? (d) Recalculate the standard error of the mean with correction for finite population. Find the 95% confidence interval for the average employee salary.

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Answer to a math question 1. The following is a simple random sample taken from a population of employees in a company. The information that was asked of each person was the monthly salary they had last year measured in dollars. 1200 4500 1340 3100 2500 5100 2100 1300 2300 4100 3200 1200 5200 3500 2600 (a) Determine the sample average wage and the sample standard deviation of the wages of the employees. (b) Determine the standard error of the mean assuming the population is infinite. Find the 95% confidence interval for the average salary of employees. (c) Assuming that the company has a total of 50 employees. How should I change the interval reliable for the average salary of employees? (d) Recalculate the standard error of the mean with correction for finite population. Find the 95% confidence interval for the average employee salary.

$Expert avatar$

Tiffany

4.5

103 Answers

(a)

\bar{X} = \frac{\sum X_i}{n} = \frac{44000}{15} = 2933.33 \, \text{dólares}

s = \sqrt{\frac{\sum (X_i - \bar{X})^2}{n-1}} = \sqrt{\frac{25100000}{14}} = 1341.94 \, \text{dólares}

(b)

E_m = \frac{s}{\sqrt{n}} = \frac{1341.94}{\sqrt{15}} = 346.53 \, \text{dólares}

\text{IC}_{95\%} = \bar{X} \pm Z \frac{s}{\sqrt{n}} = 2933.33 \pm 1.96 \cdot 346.53

\text{IC}_{95\%} = (2254.14, 3612.52) \, \text{dólares}

(c)
Dado que la población es finita (50 empleados), el intervalo de confianza debería ser más pequeño debido al factor de corrección de población finita.

(d)

E = \frac{s}{\sqrt{n}} \sqrt{\frac{N-n}{N-1}} = \frac{1341.94}{\sqrt{15}} \sqrt{\frac{50-15}{50-1}} = 316.25 \, \text{dólares}

\text{IC}_{95\%} = \bar{X} \pm Z \cdot E = 2933.33 \pm 1.96 \cdot 316.25

\text{IC}_{95\%} = (2304.21, 3562.46) \, \text{dólares}

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