Respuesta :
Answer:
The minimum sample size needed 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.
Assuming that the standard deviation of the population of systolic blood pressures of executives of major corporations is 25 mm Hg.
This means that [tex]\sigma = 25[/tex]
What is the minimum sample size needed for the researcher to be 90% confident that his estimate is within 5 mm Hg of u?
This minimum sample size is n.
n is found when M = 5. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]5 = 1.645\frac{25}{\sqrt{n}}[/tex]
[tex]5\sqrt{n} = 1.645*25[/tex]
Dividing both sides by 5
[tex]\sqrt{n} = 1.645*5[/tex]
[tex](\sqrt{n})^2 = (1.645*5)^2[/tex]
[tex]n = 67.65[/tex]
Rounding up:
The minimum sample size needed is 68.
