Question

4. The monthly incomes from a random sample of workers in a factory are shown below. Monthly Income (In $1,000) 4.0 5.0 7.0 4.0 6.0 6.0 7.0 9.0 a. Compute the standard error of the mean (in dollars). b. Compute the margin of error in dollars) at 95% confidence. c. Complute a 95% confidence interval for the mean of the population. Assume the population has a normal distribution. Give your answer in dollars.

Answer 1

Answer:-

Given that:-

monthly income in $1000

4.0, 5.0, 7.0, 4.0 , 6.0, 6.0, 7.0, 9.0

..n = 8

mean $=\overline{X}=\frac{1}{n}\sum X_{i}=\frac{1}{8}[4+5+....+7+9]$

48 X = 6. 8

S = Χ? - ? = a: [42 + 52 + ... + 92] – 62 n
$=\sqrt{\frac{308}{8}-36}=\sqrt{2.5}$

$S=1.58$ (in $1000)

$\therefore$(a) standard of mean $=\frac{S}{\sqrt{n}}$

$=\frac{1.58}{\sqrt{8}}$

$=0.5586$ ($1000)

Standard error of mean = 558.6 $

b)
g= 0,05
  $t_{n-1,1-\alpha/2}=t_{7,0.975}=2.36$

$\therefore$Margin of error, $E= t_{n-1,1-\alpha/2}\times \frac{S}{\sqrt{n}}$

$=2.36\times 0.5586$

$E=1.318$ ($1000)

c) 95% Confidence interval is given by

$(\overline{X}\pm t_{n-1,1-\alpha/2}\times \frac{S}{\sqrt{n}})$

$\equiv (6\pm 1.318)$

$\equiv (4.682,7.318)$ (in $1000)

$\equiv (4682,7318)$ in $

$\therefore$95% C.I for the mean of population is ($4,682 , $ 7,318)