Answer:
a) Confidence interval is (182.86,203.54).
b) (i) Yes, 200 bips is a true mean as it lie in the interval.
(ii) No, 210 bips is not a true mean as it doesn't lie in the interval.
Step-by-step explanation:
Given : A researcher studied the radioactivity of asbestos. She sampled 81 boards of asbestos, and found a sample mean of 193.2 bips, and a sample standard deviation of 49.5 bips.
To find : (a) Obtain the 94% confidence interval for the mean radioactivity. (b) (i) According the interval that you got, is 200 bips a plausible value for the true mean? (ii) What about 210 bips?
Solution :
a) The confidence interval formula is given by,
![\bar{x}-z\times \frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=%5Cbar%7Bx%7D-z%5Ctimes%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%20%3CC.I%3C%5Cbar%7Bx%7D%2Bz%5Ctimes%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
We have given,
The sample mean
bips
s is the standard deviation
bips
n is the number of sample n=81
z is the score value, at 94% z=1.88
Substitute all the values in the formula,
![193.2-1.88\times \frac{49.5}{\sqrt{81}}](https://tex.z-dn.net/?f=193.2-1.88%5Ctimes%20%5Cfrac%7B49.5%7D%7B%5Csqrt%7B81%7D%7D%20%3CC.I%3C193.2%2B1.88%5Ctimes%20%5Cfrac%7B49.5%7D%7B%5Csqrt%7B81%7D%7D)
![193.2-1.88\times 5.5](https://tex.z-dn.net/?f=193.2-1.88%5Ctimes%205.5%20%3CC.I%3C193.2%2B1.88%5Ctimes%205.5)
![193.2-10.34](https://tex.z-dn.net/?f=193.2-10.34%20%3CC.I%3C193.2%2B10.34)
![182.86](https://tex.z-dn.net/?f=182.86%20%3CC.I%3C203.54)
Confidence interval is (182.86,203.54).
b) (i) According the interval ![182.86](https://tex.z-dn.net/?f=182.86%20%3CC.I%3C203.54)
200 bips a plausible value for the true mean as it lies in the interval.
(ii) 210 bips not lie in the confidence interval so it is not a true mean.