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...compute a 95% confidence interval from a random sample of 10 scores: 320, 380, 400, 420, 500, 520, 600, 660, 720, and 780. Assume that the standard deviation of SAT verbal scores in a school system is known to be 100.

Also, how large should n be so that a 95% confidence interval for m has a margin of error of 0.0001?

When constructing confidence intervals, the z-distribution is used when the *population* standard deviation is known.

- In Excel, m =AVERAGE(320, 380, 400, 420, 500, 520, 600, 660, 720, 780) = 530
- Standard error of the mean = 100 / 10^0.5 = 31.62
- z = 1.96 (from the z-table)
- It turns out that one must go 1.96 standard deviations from the mean in both directions to contain .95 of the scores.

- Confidence interval = Point Estimate +/- Margin of Error = Point Estimate +/- Reliability Factor * Standard Error
- Lower limit = 530 - 1.96 * 31.62 = 468.02
- Upper limit = 530 + 1.96 * 31.62 = 591.98

- The research can be 95% certain that the mean SAT in the school system is between 468 and 592.
- To have a margin of error of 0.1: Reliability Factor * Standard Error = z * stdev /n^0.5 = 0.1
- n = (z * stdev / 0.1)^2 = (1.96 * 100 / 0.1) ^ 2 = 3841600

*Bonus Points*

- When use z-distribution to construct confidence intervals, one assumes
- normal distribution
- the population standard deviation is known.
- Scores are sampled randomly and are independent

- The value of z for the 95% confidence interval is the number of standard deviations one must go from the mean (in both directions) to contain 95% of the scores
- If a larger sample size had been used, the range of scores would have been smaller.
- The computation of the 99% confidence interval is exactly the same except that a different z value is used. The 99% confidence interval is even
*wider*than the 95% confidence interval.

*Category: Quantitative Analysis*

*Category: C++ Quant > Finance*

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