Question

Shards of clay vessels were put together to reconstruct rim diameters of the original ceramic vessels found at the Wind Mountain archaeological site. A random sample of ceramic vessels gave the following rim diameters (in centimeters):

15.9 - 13.4 - 22.1 - 12.7 - 13.1 - 19.6 - 11.7 - 13.5 - 17.7 - 18.1

(A) With mean and sample deviation verify that x bar is about equal to 15.8 cm and s equal to 3.5 cm.

(B) Compute an 80% confidence interval for the population mean (mhu) of rim diameters for such ceramic vesselsfound at the Wind Mountain archaelogical site.

Answer 1

a) In order to calculate the mean and the sample deviation we can use the following formulas:

`\bar X= \sum_(i=1)^n (x_i)/(n)` (2)

`s=\sqrt{(\sum_(i=1)^n (x_i-\bar X)^2)/(n-1)}` (3)

The mean calculated for this case is

The sample deviation calculated
`s=3.46 \approx 3.5`

b)
`15.8-1.383(3.5)/(√(10))=14.27`

`15.8+1.383(3.5)/(√(10))=17.33`

So on this case the 80% confidence interval would be given by (14.27;17.33)

Explanation:

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` represent the sample mean for the sample

`\mu` population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Part a

The confidence interval for the mean is given by the following formula:

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

In order to calculate the mean and the sample deviation we can use the following formulas:

`\bar X= \sum_(i=1)^n (x_i)/(n)` (2)

`s=\sqrt{(\sum_(i=1)^n (x_i-\bar X)^2)/(n-1)}` (3)

The mean calculated for this case is

The sample deviation calculated
`s=3.46 \approx 3.5`

Part b

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

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

Since the Confidence is 0.80 or 80%, the value of
`\alpha=0.2` and
`\alpha/2 =0.1`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.1,9)".And we see that
`t_(\alpha/2)=1.383`

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

`15.8-1.383(3.5)/(√(10))=14.27`

`15.8+1.383(3.5)/(√(10))=17.33`

So on this case the 80% confidence interval would be given by (14.27;17.33)