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. Please use student’s t to find the confidence interval at the 95% confidence level. 2) If one wants the confidence interval to be two thirds of the previous one, how many times should a student repeat? (Assuming the standard deviation is the same as the previous one)?

Answer 1

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`

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