A study by the department of education of a certain state was trying to determine the mean SAT scores of the graduating high school seniors. The study examined the scores of a random sample of 121 graduating seniors and found the mean score to be 525 with a standard deviation of 97. At the 95% confidence level, find the margin of error for the mean, rounding to the nearest tenth. (Do not write \pm±).
To find the margin of error for the mean at a 95% confidence level, you need to use a confidence interval. The margin of error represents the range above and below the sample mean within which the population mean is likely to fall.
The formula for the margin of error is:
Margin of Error = Critical Value * (Standard Deviation / Square Root of Sample Size)
First, let's find the critical value. At a 95% confidence level, we need to find the z-score associated with a 95% confidence level. This value can be found using a standard normal distribution table or a statistical calculator. The z-score for a 95% confidence level is approximately 1.96.
Next, we need to calculate the margin of error using the formula mentioned earlier:
Margin of Error = 1.96 * (97 / Square Root of 121)
Since the sample size is 121, the square root of 121 is 11.
Margin of Error = 1.96 * (97 / 11) = 17.0909090909...
Rounding to the nearest tenth, the margin of error is approximately 17.1.
Therefore, the margin of error for the mean at a 95% confidence level is 17.1.