SOLUTION: The mean weight of 15 randomly selected newborn babies at a local hospital is 7.46 lbs and the standard deviation is 0.2 lbs. Assume the weight of newborn babies has approximately

Question 1181388: The mean weight of 15 randomly selected newborn babies at a local hospital is 7.46 lbs and the standard deviation is 0.2 lbs. Assume the weight of newborn babies has approximately normal distribution.
(a) Find the margin of error for the 95% confidence interval for the mean weight of all newborn babies at this hospital. (Round your answer to two decimal places.)
(b) Use information from part (a) to fill in the banks in the following sentence:
______% of all samples of size ______ have sample means within ________ lbs of the population mean.
(c) Find a 95% confidence interval for the mean weight of all newborn babies at this hospital. (Round your answer to two decimal places.)
_______ lbs < μ < ________ lbs
(d) If you increase the confidence level ( 1−α ), will the confidence interval estimate be wider or narrower? Explain.
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's the solution:
**(a) Margin of Error:**
Since the sample size is small (n=15) and the population standard deviation is unknown, we use the t-distribution. The formula for the margin of error is:
Margin of Error = t * (s / √n)
Where:
* t is the t-score for the desired confidence level and degrees of freedom (df = n - 1 = 15 - 1 = 14)
* s is the sample standard deviation
* n is the sample size
For a 95% confidence level and df = 14, the t-score is approximately 2.145 (you'll find this using a t-table or calculator).
Margin of Error = 2.145 * (0.2 / √15)
Margin of Error = 2.145 * (0.2 / 3.873)
Margin of Error ≈ 2.145 * 0.0516
Margin of Error ≈ 0.11
The margin of error is approximately **0.11 lbs**.
**(b) Interpretation:**
95% of all samples of size 15 have sample means within **0.11** lbs of the population mean.
**(c) 95% Confidence Interval:**
The confidence interval is calculated as:
(x̄ - Margin of Error, x̄ + Margin of Error)
Where x̄ is the sample mean.
(7.46 - 0.11, 7.46 + 0.11)
(7.35, 7.57)
The 95% confidence interval is approximately **7.35 lbs < μ < 7.57 lbs**.
**(d) Effect of Increasing Confidence Level:**
If you increase the confidence level (1 - α), the confidence interval estimate will be *wider*.
*Explanation:* A higher confidence level means you want to be more certain that your interval contains the true population mean. To be more certain, you need to make the interval wider. A wider interval is more likely to capture the true mean, but it also provides less precise information about the mean's specific value. Think of it like casting a wider net to catch a fish - you're more likely to catch it, but you'll have less idea about exactly where it is.

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