Answer:
a) The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams
b) 49 measurements are needed.
Step-by-step explanation:
Question a:
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
![\alpha = \frac{1 - 0.98}{2} = 0.01](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B1%20-%200.98%7D%7B2%7D%20%3D%200.01)
Now, we have to find z in the Ztable as such z has a pvalue of
.
That is z with a pvalue of
, so Z = 2.327.
Now, find the margin of error M as such
![M = z\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In which
is the standard deviation of the population and n is the size of the sample.
![M = 2.327\frac{0.0003}{\sqrt{5}} = 0.00031](https://tex.z-dn.net/?f=M%20%3D%202.327%5Cfrac%7B0.0003%7D%7B%5Csqrt%7B5%7D%7D%20%3D%200.00031)
The lower end of the interval is the sample mean subtracted by M. So it is 10.0044 - 0.00031 = 10.00409 grams
The upper end of the interval is the sample mean added to M. So it is 10 + 0.00031 = 10.00471 grams
The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams.
(b) How many measurements must be averaged to get a margin of error of +/- 0.0001 with 98% confidence?
We have to find n for which M = 0.0001. So
![M = z\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
![0.0001 = 2.327\frac{0.0003}{\sqrt{n}}](https://tex.z-dn.net/?f=0.0001%20%3D%202.327%5Cfrac%7B0.0003%7D%7B%5Csqrt%7Bn%7D%7D)
![0.0001\sqrt{n} = 2.327*0.0003](https://tex.z-dn.net/?f=0.0001%5Csqrt%7Bn%7D%20%3D%202.327%2A0.0003)
![\sqrt{n} = \frac{2.327*0.0003}{0.0001}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B2.327%2A0.0003%7D%7B0.0001%7D)
![(\sqrt{n})^2 = (\frac{2.327*0.0003}{0.0001})^2](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E2%20%3D%20%28%5Cfrac%7B2.327%2A0.0003%7D%7B0.0001%7D%29%5E2)
![n = 48.73](https://tex.z-dn.net/?f=n%20%3D%2048.73)
Rounding up
49 measurements are needed.