Answer :
Answer:
[tex]1.39-2.262\frac{0.78}{\sqrt{10}}=0.832[/tex]
[tex]1.39+2.262\frac{0.78}{\sqrt{10}}=1.948[/tex]
Step-by-step explanation:
Assuming the following question: Compute a 95% confidence interval using the 2-distribution, as the population standard deviation is known. The bounds of this confidence interval (lower upper) are:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]\bar X =1.39[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s=0.78 represent the sample standard deviation
n=10 represent the sample size
Calculate the confidence interval
Since the sample size is large enough n<30. The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
The degrees of freedom are given by:
[tex] df =n-1= 10-1=9[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,9)".And we see that [tex]t_{\alpha/2}=2.262[/tex]
Now we have everything in order to replace into formula (1):
[tex]1.39-2.262\frac{0.78}{\sqrt{10}}=0.832[/tex]
[tex]1.39+2.262\frac{0.78}{\sqrt{10}}=1.948[/tex]
Question:
The question is incomplete. Find below the complete question and the answer.
A sample of 10 chunks of coal from a particular mine found a sample mean density of 1.39 g/m3 with a population standard deviation of 0.78 g/m3. We wish to create a confidence interval for the true mean density of coal from this particular mine. Round all answers to three decimal places.
(a) Compute a 95% confidence interval using the z-distribution.
(b)Using a critical value of 2, create a 95% confidence interval.
(c) Suppose that the population standard deviation was actually a sample standard deviation, compute a 95% confidence interval for the true mean coal density using the t-distribution.
Answer;
(a) 0.827 ∠ μ ∠ 1.893
(b) (0.816, 1.904)
(c) 0.745 ∠ μ ∠ 1.975
Step-by-step explanation:
Given data;
Sample mean density = 1.39g/m^3
Number of chucks (n) = 10
Standard deviation = 0.86g/m^3
See the attached file for explanation
