Question

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. (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?

Answer 1

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,

$\bar{x}-z\times \frac{\sigma}{\sqrt{n}} <C.I<\bar{x}+z\times \frac{\sigma}{\sqrt{n}}$

We have given,            

The sample mean $\bar{x}=193.2$ bips

s is the standard deviation $\sigma=49.5$ 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,

$193.2-1.88\times \frac{49.5}{\sqrt{81}} <C.I<193.2+1.88\times \frac{49.5}{\sqrt{81}}$

$193.2-1.88\times 5.5 <C.I<193.2+1.88\times 5.5$

$193.2-10.34 <C.I<193.2+10.34$

$182.86 <C.I<203.54$

Confidence interval is (182.86,203.54).    

b) (i) According the interval $182.86 <C.I<203.54$

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.