Question

Babies weighing less than 5.5 pounds at birth are considered "low-birth-weight babies." In the United States, 7.6% of newborns are low-birth-weight babies. The following information was accumulated from samples of new births taken from two counties.

Hamilton Shelby
Sample size 150-200
Number of low-birth-weight babies 18 22
Develop a 95% confidence interval estimate for the difference between the proportions of low-birth-weight babies in the two counties.
Is there conclusive evidence that one of the proportions is significantly more than the other? If yes, which county? Explain, using the results of part (a). Do not perform any test. Give detailed answer

Answer 1

To develop a 95% confidence interval estimate for the difference between the proportions of low-birth-weight babies in the two counties, you can use the formula: 95% CI = (p1 - p2) ± (Z * √[(p1 * (1 - p1) / n1) + (p2 * (1 - p2) / n2) )].

Step-by-step explanation:

To develop a 95% confidence interval estimate for the difference between the proportions of low-birth-weight babies in the two counties, you can use the formula:

95% CI = (p1 - p2) ± (Z * √[(p1 * (1 - p1) / n1) + (p2 * (1 - p2) / n2) )]

where p1 and p2 are the proportions of low-birth-weight babies in Hamilton and Shelby counties respectively, n1 and n2 are the sample sizes of the two counties, and Z is the z-score corresponding to the desired level of confidence (for a 95% confidence interval, Z = 1.96).

Plugging in the values for Hamilton County (p1 = 18/150 and n1 = 150) and Shelby County (p2 = 22/200 and n2 = 200), you can calculate the 95% confidence interval estimate for the difference between the proportions of low-birth-weight babies in the two counties.