Respuesta :
Answer:
A sample size of 128 is needed.
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{1-0.93}{2} = 0.035[/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]1-0.035 = 0.965[/tex], so [tex]z = 1.81[/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(square root of the variance) and n is the size of the sample.
How large should a sample be if the margin of error is 1 minute for a 93% confidence interval
We need a sample size of n, which is found when [tex]M = 1[/tex]. We have that [tex]\sigma = \sqrt{39}[/tex]. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]1 = 1.81*\frac{\sqrt{39}}{\sqrt{n}}[/tex]
[tex]\sqrt{n} = 1.81\sqrt{39}[/tex]
[tex](\sqrt{n})^2 = (1.81\sqrt{39})^2[/tex]
[tex]n = 127.8[/tex]
Rounding up
A sample size of 128 is needed.
