An experimental egg farm is raising chickens to produce low cholesterol eggs. A lab tested 12 randomly selected eggs and found that the mean amount of cholesterol was x = 186 mg. The sample standard deviation was found to be s = 19.0 mg on this group. Assume that the population is normally distributed.
Question: Find the margin of error for a 95% confidence interval. Round your answer to the nearest tenths.
Answer: 1.96(19/√16) = 9.31
Question: Find a 95% confidence interval for the mean µ cholesterol content for all experimental eggs. Assume that the population is normally distributed.
Answer:
1.96 – 9.31 = 7.35
1.96 + 9.31 = 11.27
Why is it the "√16" when n - 1 = 11?
Standard error of the mean = Standard deviation/√(n-1)
95% confidence interval = mean ± 1.96 SE. Why are you not using mean = 186?
To find the margin of error for a 95% confidence interval, we need to use the formula:
Margin of Error = Critical Value * (Sample Standard Deviation / √n)
In this case, the critical value can be found using a standard normal distribution table or calculator. For a 95% confidence level, the critical value is typically 1.96.
So, Margin of Error = 1.96 * (19.0 / √12) = 9.31 (rounded to the nearest tenths)
This means that we have a margin of error of approximately 9.31 mg for our estimate of the mean cholesterol content.
Now, to find the 95% confidence interval for the mean µ cholesterol content, we can use the formula:
Confidence Interval = Sample Mean ± Margin of Error
Using the given information, the sample mean x = 186 mg and the margin of error is 9.31 mg.
So, the lower bound of the confidence interval is 186 - 9.31 = 7.35 mg (rounded to the nearest tenths).
And the upper bound of the confidence interval is 186 + 9.31 = 11.27 mg (rounded to the nearest tenths).
Therefore, we have a 95% confidence interval for the mean cholesterol content for all experimental eggs of approximately 7.35 mg to 11.27 mg.