# A random sample of 1000 high schools students grains an average of 22 points in their second attempt at the SAT mathematics exam. The change in score has a normal distribution with standard...

A random sample of 1000 high schools students grains an average of 22 points in their second attempt at the SAT mathematics exam. The change in score has a normal distribution with standard deviation 50.

(a) Give a 95% confidence interval for the mean score gain in the population of all students. Use the 4-step method for confidence intervals.

(b) Scores from how many students must be averaged to get a margin of error of ±3 with 90% confidence?

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We are given a sample of size 1000 with a sample mean of `bar(x)=22 ` and a population standard deviation of `sigma=50 ` .

(a) We are asked to find a 95% confidence interval for the population mean.

-- For a 95% confidence interval we have `alpha=.05 ` and `z_(alpha/2)=1.96 `

( 2.5% or .025 of the population is above the interval and .025 is below. Use a standard normal distribution table to find that this corresponds to z=1.96)

-- The standard error is `sigma/sqrt(n)=50/sqrt(1000)~~1.581 `

The interval is given by `bar(x)-z_(alpha/2)(sigma/sqrt(n))<mu<bar(x)+z_(alpha/2)(sigma/sqrt(n)) `

Substituting the given values we get:

`22-1.96(50/sqrt(1000))<mu<22+1.96(50/sqrt(1000)) ` or

`18.901<mu<25.099 `

(b) If we want the mean within a margin of error of 3% with a 90% confidence:

The formula for the error is `E=z_(alpha/2)(sigma/sqrt(n)) `

Solving for n we get `n=((z_(alpha/2)*sigma)/E)^2 ` Substituting the given values we get:

`n=((1.645*50)/3)^2~~751.67 ` (Since we have a 90% confidence, the z score associated with .9500 is 1.645.)

Thus we need a minimum of 752 in our sample.

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