Answer:
The minimum sample size that should be taken is 62.
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].
So it is z with a pvalue of [tex]1-0.05 = 0.95[/tex], so [tex]z = 1.645[/tex]
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.
If we want to be 90% confident that the sample mean is within 1 word per minute of the true population mean, what is the minimum sample size that should be taken
This is n when [tex]M = 1, \sigma = 4[/tex]. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]1 = 1.96*\frac{4}{\sqrt{n}}[/tex]
[tex]\sqrt{n} = 4*1.96[/tex]
[tex](\sqrt{n})^{2} = (4*1.96)^{2}[/tex]
[tex]n = 61.5[/tex]
The minimum sample size that should be taken is 62.