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30 December, 23:19

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 81 graduating seniors and found the mean score to be 506 with a standard deviation of 83. At the 95% confidence level, find the margin of error for the mean, rounding to the nearest tenth. (Do not write ±).

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  1. 30 December, 23:41
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    18.1

    Step-by-step explanation:

    Margin of error = critical value * standard error

    At 95% confidence, the critical value is z = 1.960.

    The standard error is σ/√n = 83/√81 = 9.222.

    So the margin of error is:

    1.960 * 9.222 = 18.1
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