Respuesta :
Answer:
The smallest sample size needed is 49.
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.
Find the smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income.
This is n for which [tex]M = 500, \sigma = 2119[/tex]
So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]500 = 1.645*\frac{2119}{\sqrt{n}}[/tex]
[tex]500\sqrt{n} = 1.645*2119[/tex]
[tex]\sqrt{n} = \frac{1.645*2119}{500}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.645*2119}{500})^{2}[/tex]
[tex]n = 48.6[/tex]
Rounding up
The smallest sample size needed is 49.
Answer:
[tex]n=(\frac{1.640(2119)}{500})^2 =48.31 \approx 49[/tex]
So the answer for this case would be n=49 rounded up to the nearest integer
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]\bar X=37500[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma=2119[/tex] represent the sample standard deviation
n represent the sample size
Solution to the problem
The margin of error is given by this formula:
[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (a)
And on this case we have that ME =500 and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (b)
The critical value for 90% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.05;0;1)", and we got [tex]z_{\alpha/2}=1.640[/tex], replacing into formula (b) we got:
[tex]n=(\frac{1.640(2119)}{500})^2 =48.31 \approx 49[/tex]
So the answer for this case would be n=49 rounded up to the nearest integer
