Question

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.

Answer 1

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

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

Answer 2

`1.39-2.262(0.78)/(√(10))=0.832`

`1.39+2.262(0.78)/(√(10))=1.948`

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".

`\bar X =1.39` represent the sample mean

`\mu` 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:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

The degrees of freedom are given by:

`df =n-1= 10-1=9`

Since the Confidence is 0.95 or 95%, the value of
`\alpha=0.05` and
`\alpha/2 =0.025`, 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
`t_(\alpha/2)=2.262`

Now we have everything in order to replace into formula (1):

`1.39-2.262(0.78)/(√(10))=0.832`

`1.39+2.262(0.78)/(√(10))=1.948`