Respuesta :
Answer:
We need to sample at least 1069 parents.
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.
How many parents do you have to sample?
We need to sample at least n parents.
n is found when [tex]M = 4, \sigma = 79.5[/tex]. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]4 = 1.645*\frac{79.5}{\sqrt{n}}[/tex]
[tex]4\sqrt{n} = 1.645*79.5[/tex]
[tex]\sqrt{n} = \frac{1.645*79.5}{4}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.645*79.5}{4})^{2}[/tex]
[tex]n = 1068.92[/tex]
Rounding up
We need to sample at least 1069 parents.
Answer:
[tex]n=(\frac{1.64(79.5)}{4})^2 =1062.43 \approx 1063[/tex]
So the answer for this case would be n=1063 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[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma=79.5[/tex] represent the population 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{s}{\sqrt{n}}[/tex] (a)
And on this case we have that ME =4 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 formula to find it:"=-NORM.INV(0.05;0;1)", and we got [tex]z_{\alpha/2}=1.64[/tex], replacing into formula (b) we got:
[tex]n=(\frac{1.64(79.5)}{4})^2 =1062.43 \approx 1063[/tex]
So the answer for this case would be n=1063 rounded up to the nearest integer
