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 208 graduating seniors and found the mean score to be 525 with a standard deviation of 115. At the 95% confidence level, use the normal distribution/empirical rule to estimate the margin of error for the mean, rounding to the nearest tenth.
Question
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 208 graduating seniors and found the mean score to be 525 with a standard deviation of 115. At the 95% confidence level, use the normal distribution/empirical rule to estimate the margin of error for the mean, rounding to the nearest tenth.
Solution
To estimate the margin of error for the mean at the 95% confidence level, we will use the formula for the margin of error (E) in a normal distribution, which is:
E = Z * (σ/√n)
Where:
- Z is the Z-score, which corresponds to the desired confidence level. For a 95% confidence level, the Z-score is approximately 1.96 (from Z-table or standard normal distribution table).
- σ is the standard deviation, which is given as 115.
- n is the sample size, which is given as 208.
Substituting the given values into the formula, we get:
E = 1.96 * (115/√208)
Calculating the above expression will give us the margin of error.
Let's calculate it:
E = 1.96 * (115/√208) = 1.96 * (115/14.4222) = 1.96 * 7.975 = 15.63
Rounding to the nearest tenth, the margin of error for the mean at the 95% confidence level is approximately 15.6.
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