Someone help meh please:)simplify the complex fraction

Ответ:

Travon1418

17.08.2022

We need an SRS of scores of at least 153.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of 1-0.05 = 0.95 , so z = 1.645

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ 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 $M = 20, \sigma = 150$ . So

$M = z*\frac{\sigma}{\sqrt{n}}$

$20 = 1.645*\frac{150}{\sqrt{n}}$

$20\sqrt{n} = 1.645*150$

$\sqrt{n} = \frac{1.645*150}{20}$

$\sqrt{n} = 12.3375$

$(\sqrt{n})^{2} = (12.3375)^{2}$

n = 152.2

Rounding to the next whole number, 153

We need an SRS of scores of at least 153.

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