Question

A large school district in Connecticut wants to estimate the average SAT score of this year’s graduating class using a confidence interval. The district takes a simple random sample of 100 seniors. The sample mean is 510 and the population standard deviation for SAT scores is known to be 75. They want an interval that "likely" contains the true mean, where "likely" means "the procedure has a 68% success rate". What is the margin of error for this confidence interval?

Answer 1

Answer:

The margin of error (M.E) = 3.51

Step-by-step explanation:

Explanation:-

Given data a large school district in Connecticut wants to estimate the average SAT score of this year’s graduating class.

The district takes a simple random sample of 100 seniors.

The sample size  'n' = 100

given data the sample mean is 510 and the population standard deviation for SAT scores is known to be 75.

mean of the sample  = 510

population standard deviation σ = 75.

Given data they want an interval that "likely" contains the true mean, where "likely" means "the procedure has a 68% success rate".

level of significance = 68% =0.68

Z-score of 68% of level of significance = 0.468

Margin of error:-

The margin of error of at 68% of confidence intervals.

$M.E = \frac{Z_{0.68} S.D }{\sqrt{n} }$

$M.E = \frac{0.468 X 75 }{\sqrt{100} } = 3.51$

Margin of error = 3.51

Conclusion:-

Margin of error = 3.51 at 68% of confidence interval