Mathematics: Achemist made eight independent measurements of the melting point of tungsten. she obtained a sample mean of 3410.14 degrees celsius and a sample standard deviation of 1.018 degrees. a) use the student’s t distribution to find a 95% confidence interval for the melting point of tungsten. b) use the student’s t distribution to find a 98% confidence interval for the melting point of tungsten.

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Achemist made eight independent measurements of

Ответ:

therealpr1metime45

13.01.2022

Mathematics

a) The 95% confidence interval would be given by (3409.291;3410.989)

b) The 98% confidence interval would be given by (3409.060;3411.220)

Step-by-step explanation:

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".

Part a

$\bar X=3410.14$ represent the sample mean

$\mu$ population mean (variable of interest)

s=1.018 represent the sample standard deviation

n=8 represent the sample size

95% confidence interval

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

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

First we need to find the degrees of freedom given by:

df=n-1=8-1=7

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(0.025,7)".And we see that $t_{\alpha/2}=2.36$

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

$3410.14-2.36\frac{1.018}{\sqrt{8}}=3409.291$

$3410.14+2.36\frac{1.018}{\sqrt{8}}=3410.989$

So on this case the 95% confidence interval would be given by (3409.291;3410.989)

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(0.01,7)".And we see that $t_{\alpha/2}=3.00$

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

$3410.14-3.00\frac{1.018}{\sqrt{8}}=3409.060$

$3410.14+3.00\frac{1.018}{\sqrt{8}}=3411.220$

So on this case the 98% confidence interval would be given by (3409.060;3411.220)

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