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 130 graduating seniors and found the mean score to be 458 with a standard deviation of 105. 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 130 graduating seniors and found the mean score to be 458 with a standard deviation of 105. 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 calculate 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 105.
- n is the sample size, which is given as 130.
Substituting the given values into the formula, we get:
E = 1.96 * (105/√130)
Calculating the above expression will give us the margin of error.
Let's calculate it:
E = 1.96 * (105/√130) = 1.96 * (105/11.4) = 1.96 * 9.21 = 18.05
Rounding to the nearest tenth, the margin of error for the mean at the 95% confidence level is approximately 18.1.
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