Respuesta :
Answer:
We need at least 4238 scores.
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{1-0.97}{2} = 0.015[/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]1-0.015 = 0.985[/tex], so [tex]z = 2.17[/tex]
Now, find the margin of error M
[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 random IQ scores must be obtained if we want to find the true population mean (with an allowable error of 0.5) and we want 97 percent confidence in the results
We need at least n scores, in which N is found when [tex]M = 0.5[/tex]. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.5 = 2.17*\frac{15}{\sqrt{n}}[/tex]
[tex]0.5\sqrt{n} = 2.17*15[/tex]
[tex]0.5\sqrt{n} = 32.55[/tex]
[tex]\sqrt{n} = \frac{32.55}{0.5}[/tex]
[tex]\sqrt{n} = 65.1[/tex]
[tex]\sqrt{n}^{2} = (65.1)^{2}[/tex]
[tex]n = 4238[/tex]
We need at least 4238 scores.
