Answer:
So, the required sample size is
, that she will need to be 95% confident that her sample mean will be within 15 seconds of the true mean.
Step-by-step explanation:
Given that,
Standard deviation (
) = 40
and we have to find how large a sample is needed to be 95% confident that her sample mean will be within 15 seconds of the true mean.
Now,
If
is used as an estimate of
, we can be
% confident that the error will not exceed a specified amount
when the sample size is
........ (i)
where
is the z value leaving an area of
to the right.
So,
% = 95%
![1-\alpha=0.95](https://tex.z-dn.net/?f=1-%5Calpha%3D0.95)
![\alpha = 0.05](https://tex.z-dn.net/?f=%5Calpha%20%3D%200.05)
![\frac{\alpha}{2} = 0.025](https://tex.z-dn.net/?f=%5Cfrac%7B%5Calpha%7D%7B2%7D%20%3D%200.025)
We have
and
, Now using equation (i), the required sample size is,
........(ii)
Now we have to find the value of
.
So, the
is the z-value leaving an area of 0.025 to the right {the area left of the
is (1-0.025) = 0.975.}
Using Normal Probability table, we see that the closest z-value which leaving an area of 0.025 to the right (i.e. an area of 0.975 to the left) is ![Z_{0.025} = 1.96](https://tex.z-dn.net/?f=Z_%7B0.025%7D%20%3D%201.96)
Now,
Using equation (ii),
![n= [\frac{Z_{0.025}\times 40 }{15}]^{2}](https://tex.z-dn.net/?f=n%3D%20%5B%5Cfrac%7BZ_%7B0.025%7D%5Ctimes%2040%20%7D%7B15%7D%5D%5E%7B2%7D)
= ![[\frac{1.96\times 40}{15}]^{2}](https://tex.z-dn.net/?f=%5B%5Cfrac%7B1.96%5Ctimes%2040%7D%7B15%7D%5D%5E%7B2%7D)
≈ ![27.3](https://tex.z-dn.net/?f=27.3)
So, the required sample size is ![28.](https://tex.z-dn.net/?f=28.)