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)?

1)    

b)

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

The margin of error is given by this formula:

   (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:

  (2)

Replacing we got:

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

Step-by-step 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:

represent the sample mean

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:

  (1)

The degrees of freedom, given by:

The Confidence level is 0.95 or 95%, and the significance would be and , the critical value would be using the t distribution with 4 degrees of freedom:

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

Part 2

The original margin of error is given by:

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

The margin of error is given by this formula:

   (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:

  (2)

Replacing we got:

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

207/8 is the right

d

step-by-step explanation: