Question

1) Lithium isotope rations are important to medicine, the 6Li/7Li ratio in a standard reference material was measured several times, and the values are: 0.082601, 0.082621, 0.082589, 0.082617, 0.082598.

asked Jul 5, 2021 8.7k views

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Monis Iqbal · Answer 1 · 2021-07-10T20:09:55+0000

Answer:

1)
0.0826052-2.776(0.000013424)/(√(5))=0.082588

0.0826052+2.776(0.000013424)/(√(5))=0.0826219

b)
ME= 2.776(0.000013424)/(√(5))=0.0000166653

And we want 2/3 of the margin of error so then would be:
2/3 ME = 0.00001111

The margin of error is given by this formula:

$ME=z_(\alpha/2)(s)/(√(n))$ (1)

And on this case we have that ME =0.00001111016 and we are interested in order to find the value of n, if we solve n from equation (1) we got:

$n=((z_(\alpha/2) s)/(ME))^2$ (2)

Replacing we got:

$n=((2.776(0.000013424))/(0.00001111))^2 =11.25 \approx 12$

So the answer for this case would be n=12 rounded up to the nearest integer

Explanation:

Information given

0.082601, 0.082621, 0.082589, 0.082617, 0.082598

We can calculate the sample mean and deviation with the following formulas:

$\bar X= (\sum_(i=1)^n X_i)/(n)$

$s = \sqrt{(\sum_(i=1)^n (X_i -\bar X)^2)/(n-1)}$

$\bar X=0.0826052$ represent the sample mean

$\mu$ population mean

s=0.000013424 represent the sample standard deviation

n=5 represent the sample size

Part 1

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, given by:

df=n-1=5-1=4

The Confidence level is 0.95 or 95%, and the significance would be
$\alpha=0.05$ and
$\alpha/2 =0.025$ , the critical value would be using the t distribution with 4 degrees of freedom:
$t_(\alpha/2)=2.776$

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

0.0826052-2.776(0.000013424)/(√(5))=0.082588

0.0826052+2.776(0.000013424)/(√(5))=0.0826219

Part 2

The original margin of error is given by:

ME= 2.776(0.000013424)/(√(5))=0.0000166653