Question

Suppose that scores on the mathematics part of a test for eighth-grade students follow a Normal distribution with standard deviation σ = 150. You want to estimate the mean score within ±20 with 90% confidence. How large an SRS of scores must you choose? (Round your answer up to the next whole number.)

Answer 1

We need an SRS of scores of at least 153.

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.

Answer 2

We need an SRS of scores of at least 153.

Explanation:

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

`\alpha = (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*(\sigma)/(√(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*(\sigma)/(√(n))`

`20 = 1.645*(150)/(√(n))`

`20√(n) = 1.645*150`

`√(n) = (1.645*150)/(20)`

`√(n) = 12.3375`

`(√(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.