Question

A chemist made six independent measurements of the sublimation point of carbon dioxide (the temperature at which it changes to dry ice). She obtained a sample mean of 196.64 K with a standard deviation of 0.66 K. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part.

Required:
(a) Use the Student's t distribution to find a 95% confidence interval for the sublimation point of carbon dioxide.
(b) Use the Student's t distribution to find a 98% confidence interval for the sublimation point of carbon dioxide.
(c) If the six measurements had been 196.35, 196.32, 196.4 198.02, 196.36, 196.39, would the confidence intervals above be valid? Explain.

Answer 1

a)
`196.64-2.571(0.66)/(√(6))=195.947`

`196.64 +2.571(0.66)/(√(6))=197.333`

So on this case the 95% confidence interval would be given by (195.947;197.333)

b)
`196.64-3.365(0.66)/(√(6))=195.733`

`196.64 +3.365(0.66)/(√(6))=197.547`

So on this case the 98% confidence interval would be given by (195.733;197.547)

c) Since we are using the t distribution, a we assume random sampling method in order to obtain the data we can approximate the sample mean
`\bar X`, and the results are valid for this case.

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=196.64` represent the sample mean for the sample

`\mu` population mean (variable of interest)

s=0.66 represent the sample standard deviation

n=6 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 critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=6-1=5`

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(0.025,5)".And we see that
`t_(\alpha/2)=2.571`

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

`196.64-2.571(0.66)/(√(6))=195.947`

`196.64 +2.571(0.66)/(√(6))=197.333`

So on this case the 95% confidence interval would be given by (195.947;197.333)

Part b

Since the Confidence is 0.98 or 98%, the value of
`\alpha=0.02` and
`\alpha/2 =0.01`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,5)".And we see that
`t_(\alpha/2)=3.365`

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

`196.64-3.365(0.66)/(√(6))=195.733`

`196.64 +3.365(0.66)/(√(6))=197.547`

So on this case the 98% confidence interval would be given by (195.733;197.547)

Part c

Data: 196.35, 196.32, 196.4 198.02, 196.36, 196.39

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))/(n-1)}` (3)

The mean calculated for this case is
`\bar X=196.64`

The sample deviation calculated
`s=0.677`

Since we are using the t distribution, a random sampling method in order to obtain the data we can approximate the sample mean
`\bar X`, and the results are valid for this case.