Answer:
The smallest sample size that we should consider is 68.
Explanation:
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
![\alpha = (1 - 0.9)/(2) = 0.05](https://img.qammunity.org/2022/formulas/mathematics/college/6f1tjkp3rjc0m3m8s8vk053td5tlym692v.png)
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 = 1.645.
Now, find the margin of error M as such
![M = z(\sigma)/(√(n))](https://img.qammunity.org/2022/formulas/mathematics/college/p19w5m3ctzqxc0b7ic9kz7y4ab19d7zpbv.png)
In which
is the standard deviation of the population and n is the size of the sample.
Standard deviation of 10 minutes.
This means that
![\mu = 10](https://img.qammunity.org/2022/formulas/mathematics/college/ivfmp0sr7xiaramupdlaws5z8nvcwsknuo.png)
What is the smallest sample size that we should consider?
This is n for which M = 2. So
![M = z(\sigma)/(√(n))](https://img.qammunity.org/2022/formulas/mathematics/college/p19w5m3ctzqxc0b7ic9kz7y4ab19d7zpbv.png)
![2 = 1.645(10)/(√(n))](https://img.qammunity.org/2022/formulas/mathematics/college/7g31zodm60hjg6s7vvbzgsqheo2rkhlkvl.png)
![2√(n) = 1.645*10](https://img.qammunity.org/2022/formulas/mathematics/college/29aqtk7312esoobq6ftt3ullall7wzk1wd.png)
![√(n) = 1.645*5](https://img.qammunity.org/2022/formulas/mathematics/college/g8ianimadt2nqelthyp04nf0ajsg3b4o60.png)
![(√(n))^2 = (1.645*5)^2](https://img.qammunity.org/2022/formulas/mathematics/college/5qo60co0fxqr2p98jnvuu764mjgbn6nh22.png)
![n = 67.7](https://img.qammunity.org/2022/formulas/mathematics/college/6sjwncwlmuivkgumeeriie7ued5tp240o6.png)
Rounding up:
The smallest sample size that we should consider is 68.