Question

Assume a random sample of the birth weights of 186 healthy babies has a mean of 3103 g and a standard deviation of 696 g. construct a 95% confidence interval estimate of the mean weight of all healthy babies born to healthy mothers. what does the interval suggest about a recent study informing soon-to-be-parents that they can expect their new baby to weigh about 2980 g?

Answer 1

Answer:

(3054.01g ,3151.99 g)

The study underestimated the weight of new born babies as 2980 g is below the mean interval.

Step-by-step explanation:

Sample size, n= 186

Sample mean, $\bar X=3103$

Sample S.D= 696

For 95% confidence interval, the Z =1.96  (since its double tailed)

The interval can be found as;

$(\bar X-Z*\frac{S.D}{\sqrt{n} } , \bar X+ Z*\frac{S.D}{\sqrt{n} } )\\\\(3103-0.96*\frac{696}{\sqrt{186} } , 3103+ 1.96*\frac{696}{\sqrt{186} } \\\\(3054.01,3151.99)$

From this result, we can predict with 95% certainty that the mean weight of all healthy babies born to healthy mothers lies between 3054.01 g and 3151.99 g.

The study underestimated the weight of new born babies as 2980 g is below the mean interval.