Answer:
We need an SRS of scores of at least 153.
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 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 large an SRS of scores must you choose?
This is at least n, in which n is found when [tex]M = 20, \sigma = 150[/tex]. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]20 = 1.645*\frac{150}{\sqrt{n}}[/tex]
[tex]20\sqrt{n} = 1.645*150[/tex]
[tex]\sqrt{n} = \frac{1.645*150}{20}[/tex]
[tex]\sqrt{n} = 12.3375[/tex]
[tex](\sqrt{n})^{2} = (12.3375)^{2}[/tex]
[tex]n = 152.2[/tex]
Rounding to the next whole number, 153
We need an SRS of scores of at least 153.
Answer:
We need an SRS of scores of at least 153.
Step-by-step explanation:
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2.
So:Now, we have to find z in the Ztable as such z has a pvalue of .
So it is z with a pvalue of ,
so Now, find M as such
In which is the standard deviation of the population and n is the size of the sample.
How large an SRS of scores must you choose?
This is at least n, in which n is found when .
So Rounding to the next whole number, 153
We need an SRS of scores of at least 153.
