Answer:
The smallest sample size that we should consider is 68.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1 - 0.9}{2} = 0.05[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1 - \alpha[/tex].
That is z with a pvalue of [tex]1 - 0.05 = 0.95[/tex], so Z = 1.645.
Now, find the margin of error M as such
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
Standard deviation of 10 minutes.
This means that [tex]\mu = 10[/tex]
What is the smallest sample size that we should consider?
This is n for which M = 2. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]2 = 1.645\frac{10}{\sqrt{n}}[/tex]
[tex]2\sqrt{n} = 1.645*10[/tex]
[tex]\sqrt{n} = 1.645*5[/tex]
[tex](\sqrt{n})^2 = (1.645*5)^2[/tex]
[tex]n = 67.7[/tex]
Rounding up:
The smallest sample size that we should consider is 68.
