The mean weight of 20 randomly selected newborn babies at a local hospital is 7.08 lbs and the standard deviation is 0.1 lbs. Assume the weight of newborn babies has approximately normal distribution.

Question

The mean weight of 20 randomly selected newborn babies at a local hospital is 7.08 lbs and the standard deviation is 0.1 lbs. Assume the weight of newborn babies has approximately normal distribution.

(a) Find the margin of error for the 90% 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 90% confidence interval for the mean weight of all newborn babies at this hospital. (Round your answer to two decimal places.)

   lbs    lbs

(d) Does the confidence interval, at 90% level, provide sufficient evidence that the mean weight of a newborn at this hospital is above 6.1 lbs? Write the appropriate inequality to justify your answer.

 

(e) If you increase the confidence level (), will the confidence interval estimate be wider or narrower? Explain.

Answer 1

(a) To find the margin of error for the 90% confidence interval for the mean weight of all newborn babies at the hospital, we must use the formula: Margin of Error = Z-score * σ/sqrt(n). Based on the information given, we can assume that the Z-score for a 90% confidence interval is 1.645, σ = 0.1 lbs and n = 20. If we plug this data into the formula, we get the following equation: Margin of Error = 1.645 * 0.10 lbs/sqrt(20) We can simplify this equation as follows: Margin of Error = 0.0677 lbs Therefore, the margin of error for the 90% confidence interval for the mean weight of all newborns at the hospital is 0.08 lbs. (b) The sentence should read: 90% of all samples of size 20 have sample means within 0.08 lbs of the population mean. (c) To find a 90% confidence interval for the mean weight of all newborn babies at the hospital, we must use the formula: Confidence Interval = Mean ± Margin of Error Substituting in the mean (7.08 pounds) and the margin of error (0.08 lbs) we get the following equation: Confidence Interval = 7.08 ± 0.08 lbs Therefore, a 90% confidence interval for the mean weight of all newborn babies at the hospital is (6.96 - 7.20) lbs. (d) To answer whether the confidence interval, at 90% level, provides sufficient evidence that the mean weight of a newborn at this hospital is above 6.1 lbs, we must use an inequality. From our interval in part (c), we know that for our population the mean weight is greater than 6.96 lbs, which is greater than the 6.1 lbs value given in the question. Therefore, we can say that 6.96 lbs > 6.1 lbs. (e) If you increase the confidence level, the confidence interval estimate will be wider. This is because a higher confidence level means that the margin of error must be increased, thus widening the interval.