Respuesta :
Answer:
We would have to take a sample of 62 to achieve this result.
Step-by-step explanation:
Confidence level of 95%.
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 - 0.95}{2} = 0.025[/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 - 0.025 = 0.975[/tex], so Z = 1.96.
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.
Assume that the standard deviation in the amount of caffeine in 8 ounces of decaf coffee is known to be 2 mg.
This means that [tex]\sigma = 2[/tex]
If we wanted to estimate the true mean amount of caffeine in 8 ounce cups of decaf coffee to within /- 0.5 mg, how large a sample would we have to take to achieve this result?
We would need a sample of n.
n is found when [tex]M = 0.5[/tex]. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.5 = 1.96\frac{2}{\sqrt{n}}[/tex]
[tex]0.5\sqrt{n} = 2*1.96[/tex]
Dividing both sides by 0.5
[tex]\sqrt{n} = 4*1.96[/tex]
[tex](\sqrt{n})^2 = (4*1.96)^2[/tex]
[tex]n = 61.5[/tex]
Rounding up
We would have to take a sample of 62 to achieve this result.
