Here are summary statistics for randomly selected weights of newborn girls: n=217, x=31.7hg, s=6.5hg. Construct a confidence interval estimate of the mean. Use a 98% confidence level. Are these results very different from the confidence interval 29.6hg< u < 32.6hg with only 17 sample values, x=31.1hg, and s=2.4hg?
What is the confidence interval for the population mean ?
hg < u < hg
(Round to one decimal place as needed.)
To construct a confidence interval estimate of the mean, we can use the formula:
Confidence Interval = x ± (Z * (s/√n))
Where:
- x is the sample mean
- Z is the z-score associated with the desired confidence level
- s is the sample standard deviation
- n is the sample size
In this case, we are given:
- n = 217
- x = 31.7 hg
- s = 6.5 hg
- Confidence level = 98%
To find the z-score corresponding to a 98% confidence level, we need to use a standard normal distribution table or a calculator. The z-score for a 98% confidence level is approximately 2.33.
Now we can substitute the values into the formula:
Confidence Interval = 31.7 ± (2.33 * (6.5/√217))
Calculating this gives us:
Confidence Interval ≈ 31.7 ± (2.33 * 0.442)
Simplifying further, we have:
Confidence Interval ≈ 31.7 ± 1.028
Lower Confidence Limit ≈ 30.672
Upper Confidence Limit ≈ 32.728
Therefore, the confidence interval for the population mean is approximately 30.7 hg < u < 32.7 hg
To determine if these results are very different from the confidence interval obtained with 17 sample values, we need to compare the ranges of the two intervals.
The first interval with 217 sample values has a range of 32.728 - 30.672 = 2.056 hg.
The second interval with 17 sample values has a range of 32.6 - 29.6 = 3 hg.
Since the second interval has a larger range, we can conclude that the results obtained with the 17 sample values are more different than the results obtained with the 217 sample values.