With 95% confidence, the researcher estimates the mean birth weight of all babies born to women who are over the age of 40 to be between 2935 and 3135 grams. The researcher wants to maintain the 95% level of confidence but report a confidence interval with a smaller margin of error. What should she do

This question is incomplete, the complete question is;

A medical researcher wants to estimate μ, the mean weight of babies born to women over the age of 40.

The researcher chooses a random sample of 100 pregnant women who are over 40. Using the mean birth weight of the 100 babies in the sample, the researcher calculates the 95% confidence interval for μ.

With 95% confidence, the researcher estimates the mean birth weight of all babies born to women who are over the age of 40 to be between 2935 and 3135 grams.

The researcher wants to maintain the 95% level of confidence but report a confidence interval with a smaller margin of error. What should she do

a) Redo the study and choose a different sample of size 100

b) Redo the study, but this time with a sample size 64

c) Redo the study, but this time with a sample 225

Option C) is the correct answer.

C) Redo the study, but this time with a sample 225

Step-by-step explanation:

Given that;

A medical researcher wants to estimate μ, the mean weight of babies born to women over the age of 40.

n = 100

95% confidence interval is given ( 2935, 3135)

But the researcher wants to maintain the 95% level of confidence as well as smaller margin of error.

Therefore we have to reduce the margin of Error

and we know that if the sample size increases, the margin of Error decreases i.e it becomes smaller.

so the best option for the researcher is to redo the study with a larger sample size so the margin of Error becomes smaller.

Option C) is the correct answer.

c) Redo the study, but this time with a sample 225

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2