Q. 3.216 - Elementary Statistics

Question

Data on low-birth-weight babies were collected over a 2-year period by 14 participating centers of the National Institute of Child Health and Human Development Neonatal Research Network. Results were reported by J. Lemons et al. in the on-line paper "Very Low Birth Weight Outcomes of the National Institute of Child Health and Human Development Neonatal Research Network". For the 1048  surviving babies whose birth weights were 751-1000 grams, the average length of stay in the hospital was 86 days, although one center had an average of 66 days and another had an average of 108 days.

Part (a): Can the mean lengths of stay be considered population means? Explain your answers.

Part (b): Assuming that the population standard deviation is 12 days, determine the z-score for a baby's length of stay of 86 days at the center where the mean was 66 days.

Part (c): Assuming that the population standard deviation is 12 days, determine the z-score for a baby's length of stay of 86 days at the center where the mean was 108 days. 

Part (d): What could you conclude from parts (b) and (c) about an infant with a length of stay equal to the mean at all centers if that infant was born at a center with a mean of 66 days? mean of 108 days?

Step-by-Step Solution

Answer

Part (a): Yes, the means are sample means because they have taken 1084 surviving babies of birth weights.

Part (b): The z-score for a baby's length of stay of 86 days at the center where the mean was 66 days is 1.67.

Part (c): The z-score for a baby's length of stay of 86 days at the center where the mean was 108 days is -1.83.

Part (d): The origin of the mean baby's length of stay are different so the scores are different from parts (a) and (b).

Part (a) Step 1. Given information.

Consider the given question,

Total number of surviving babies is 1084 whose birth weights were 751-1000 grams.

So, the answer is yes.

Part (a) Step 2. Determine whether the mean lengths of stay be considered population means.

The means are sample means because they have taken 1084 surviving babies of birth weights.

Part (b) Step 1. Determine the z -score for a baby's length of stay where the mean was 66 days .

Mean and standard deviation are 66 and 12 respectively.

The z-score for a baby's length of stay of 86 days at the center,

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{86-66}{12} \\ =1.67 \end{gathered}$$

Part (c) Step 1. Determine the z -score for a baby's length of stay where the mean was 108 days .

Mean and standard deviation are 108 and 12 respectively.

The z-score for a baby's length of stay of 86 days at the center,

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{86-108}{12} \\ =-1.83 \end{gathered}$$

Part (c) Step 1. Conclude the results from parts (b) and (c).

Consider the parts (b) and (c),

The origin of the mean baby's length of stay are different so the scores are different.