Question

High school students who take the SAT Math exam a second time generally score higher than on their first try. Past data suggest that the score increase has a standard deviation of about 50 points. How large a sample of high school students would be needed to estimate the mean change in SAT score to within 2 points with 92% confidence?

Answer 1

To estimate the mean change in SAT score to within 2 points with 92% confidence, a sample size of 11025 high school students would be needed.

Step-by-step explanation:

To estimate the mean change in SAT score to within 2 points with 92% confidence, we need to determine the sample size required. The formula to calculate the required sample size (n) for estimating the population mean with a specific margin of error can be given as:

n = (Z^2 * σ^2) / E^2

Where:

• n is the sample size
• Z is the z-score corresponding to the desired confidence level (in this case, 92% confidence corresponds to a z-score of 1.75)
• σ is the standard deviation of the population (which is given as 50 points)
• E is the desired margin of error (which is 2 points)

Substituting the values into the formula:

n = (1.75^2 * 50^2) / 2^2 = 44100 / 4 = 11025

Therefore, a sample size of 11025 high school students would be needed to estimate the mean change in SAT score to within 2 points with 92% confidence.

Answer 2

n=1915

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`\bar X` represent the sample mean for the sample

`\mu` population mean

`\sigma=50` represent the population standard deviation (assumed)

n represent the sample size (variable of interest)

ME=2 represent the margin of error desired

Confidence =0.92 or 92%

`\alpha=0.08` represent the significance level

2) Solution to the problem

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

The margin of error is given by this formula:

`ME=z_(\alpha/2)(s)/(√(n))` (a)

And on this case we have that ME =2 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

`n=((z_(\alpha/2) s)/(ME))^2` (b)

The critical value for 92% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.04,0,1)", and we got
`z_(\alpha/2)=1.75`, replacing into formula (b) we got:

`n=((1.75(50))/(2))^2 =1914.06 \approx 1915`

So the answer for this case would be n=1915 rounded up to the nearest integer