Respuesta :
Answer:
(a) Point estimate of the population mean = 10
(b) Point estimate of the population standard deviation = 3.21
(c) The margin of error for the estimation of the population mean = 2.2
(d) 95% confidence interval for the population mean = [7.8 , 12.2] .
Step-by-step explanation:
Arranging our sample data given from a normal population in ascending order we get ; 5+ 6+ 9+ 10+ 11+ 12+ 13+ 14
(a) Point estimate of population mean is given by Xbar;
Xbar = [tex]\frac{\sum X_i}{n}[/tex] where n = 8 {number of obs. in data}
Xbar = [tex]\frac{5+ 6+ 9+ 10+ 11+ 12+ 13+ 14}{8}[/tex] = 10
(b) Point estimate of the population standard deviation is given by [tex]\sigma[/tex] ;
Standard deviation formula = [tex]\frac{\sum (X_i - Xbar)^{2}}{n-1}[/tex]
Solving above formula we get point estimate of the population standard deviation = 3.21 .
(c) Margin of error for the estimation of the population mean is given by the expression ;
95% Confidence Interval for population mean = Xbar [tex]\pm[/tex] Margin of error
where Margin of error = 1.96 * [tex]\frac{\sigma}{\sqrt{n} }[/tex] {Here 1.96 is written because at 5% level
of significance z table has a value of 1.96
for two tail}
Therefore, Margin of error = [tex]1.96*\frac{3.21}{\sqrt{8} }[/tex] = 2.2
(d) 95% Confidence Interval for population mean = Xbar [tex]\pm[/tex] 1.96 * [tex]\frac{\sigma}{\sqrt{n} }[/tex]
= 10 [tex]\pm[/tex] [tex]1.96*\frac{3.21}{\sqrt{8} }[/tex] = 10 [tex]\pm[/tex] 2.2
= [10 - 2.2 , 10 + 2.2]
Therefore, 95% C.I. for population mean = [7.8 , 12.2] .
The sample data, 10, 9, 12, 14, 13, 11, 6, 5, obtained from a normal
population gives the following values;
a. 10
b. 3.21
c. 2.2
d. 95% C.I. is; 7.8 < μ < 12.2
How can the normal population data be evaluated?
a. The point estimate is an estimate of the population mean.
The point estimate is found as follows;
(10 + 9 + 12 + 14 + 13 + 11 + 6 + 5) ÷ 8 = 10
- The point estimate of the population mean, [tex]\overline{x}[/tex] = 10
b. The point estimate of the population standard deviation is given by the sample standard deviation as follows;
Sample variance, s² = ((10 - 10)² + (9 - 10)² + (12 - 10)² + (14 - 10)² + (13 - 10)² + (11 - 10)² + (6 - 10)² + (5 - 10)²) ÷ (8 - 1) ≈ 10.286
s ≈ √(10.286) ≈ 3.21
- The point estimate for the population standard deviation, s ≈ 3.21
c. The margin of error, MOE, is given by the formula;
[tex]MOE_{\gamma} = \mathbf{z_{\gamma} \times \sqrt{\dfrac{s^2}{n} }}[/tex]
Where;
z-value at 95% confidence level = 1.96
Which gives;
[tex]MOE_{\gamma} = 1.96 \times \sqrt{\dfrac{10.286^2}{8} } \approx 2.2[/tex]
- The MOE for the estimation of the population mean, [tex]MOE_{\gamma}[/tex] ≈ 2.2
d. The 95% confidence interval for the population mean is found as follows;
[tex]\mathbf{\overline {x} \pm z_{\alpha/2} \cdot \dfrac{s}{\sqrt{n} }}[/tex]
Which gives;
[tex]10 - 1.96 \times \dfrac{3.207}{\sqrt{8} } < \mu < \mathbf{10 + 1.96 \times \dfrac{3.207}{\sqrt{8} }}[/tex]
- The 95% confidence interval is approximately 7.8 < μ < 12.2
Learn more about the normal (probability) distribution here:
https://brainly.com/question/1804405
