Mathematics, 30.03.2020 19:04 whitefraide

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.

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Answer from: csuggs8

a) $196.64-2.571\frac{0.66}{\sqrt{6}}=195.947$

$196.64 +2.571\frac{0.66}{\sqrt{6}}=197.333$

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

b) $196.64-3.365\frac{0.66}{\sqrt{6}}=195.733$

$196.64 +3.365\frac{0.66}{\sqrt{6}}=197.547$

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.

Step-by-step 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:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (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):

$196.64-2.571\frac{0.66}{\sqrt{6}}=195.947$

$196.64 +2.571\frac{0.66}{\sqrt{6}}=197.333$

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):

$196.64-3.365\frac{0.66}{\sqrt{6}}=195.733$

$196.64 +3.365\frac{0.66}{\sqrt{6}}=197.547$

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:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (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.

Answer from: Quest

A: 0.26 would be the answer

Answer from: Quest

table 3 represents a direct variation function. : )

Which table represents a direct variation function?

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A chemist made six independent measurements of the sublimation point of carbon dioxide (the temperat...

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