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 224 graduating seniors and found the mean score to be 492 with a standard deviation of 90. At the 95% confidence level, find the margin of error for the mean, rounding to the nearest tenth. (Do not write \pm±).
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 137 graduating seniors and found the mean score to be 448 with a standard deviation of 91. At the 95% confidence level, find the margin of error for the mean, rounding to the nearest tenth.
To find the margin of error for the mean at a 95% confidence level, you can use the formula:
Margin of Error = Critical Value * (Standard Deviation / √(Sample Size))
First, let's find the critical value. Since we are working with a 95% confidence level, the confidence level outside the critical region is 100% - 95% = 5%. Half of this is 2.5%, which corresponds to the area under the normal distribution curve.
Using a standard normal distribution table or calculator, we find the critical value for a 5% tail to be approximately 1.96.
Now, plug in the values into the formula:
Margin of Error = 1.96 * (90 / √224)
Calculating the square root of 224, we get √224 ≈ 14.97.
Substituting in the values:
Margin of Error ≈ 1.96 * (90 / 14.97)
Calculating the division, we get 90 / 14.97 ≈ 6.01.
Finally, multiplying 1.96 by 6.01, we get:
Margin of Error ≈ 1.96 * 6.01 ≈ 11.76
Rounding to the nearest tenth, the margin of error for the mean SAT scores is approximately 11.8.
To find the margin of error for the mean at a 95% confidence level, you can use the formula:
Margin of Error = Z * (Standard Deviation / √n)
Where:
- Z is the critical value that corresponds to the confidence level. For a 95% confidence level, the Z-value is 1.96 (which is obtained from the standard normal distribution).
- Standard Deviation is the known standard deviation of the population.
- n is the sample size.
In this case, the known standard deviation is 90, and the sample size is 224.
Substituting the values into the formula:
Margin of Error = 1.96 * (90 / √224)
First, calculate the square root of 224:
√224 ≈ 14.97
Now, substitute the values into the formula:
Margin of Error = 1.96 * (90 / 14.97)
Calculating the division:
Margin of Error ≈ 1.96 * 6.01
Finally, calculate the product:
Margin of Error ≈ 11.77
Therefore, the margin of error for the mean at a 95% confidence level is approximately 11.8 (rounded to the nearest tenth).
It's been 65 years since I last did stats problems dealing with confidence
level and margins of error, forgot how to do that
Hated stats then, hated teaching stats, and still don't like it, sorry
Hope oobleck can help you out