Answer:
b. False
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 he desires a 90 percent confidence interval estimate and wishes to have a margin of error of 1 minute, the required sample size will be
Sample size of n when [tex]M = 1, \mu = 4[/tex]. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]1 = 1.645*\frac{4}{\sqrt{n}}[/tex]
[tex]\sqrt{n} = 4*1.645[/tex]
[tex](\sqrt{n})^{2} = (4*1.645)^{2}[/tex]
[tex]n = 43.29[/tex]
Rouding up, 44.
Approximately 143.
False
