PLEASE SHOW ALL WORK We have survey data on the body mass in

PLEASE SHOW ALL WORK

We have survey data on the body mass index (BMI) of 649 young women. The mean BMI in the sample was x = 26. We treated these data as an SRS from a Normally distributed population with standard deviation a = 7. Suppose that we had an SRS of just 92 young women. What would be the margin of error for 95% confidence? (Round your answer to four decimal places.) Find the margins of error for 95% confidence based on SRSs of 413 young women and 1598 young women. (Round your answers to four decimal places.) n margin of error Compare the three margins of error. How does increasing the sample size change the margin of error of a confidence interval when the confidence level and population standard deviation remain the same? Margin of error remains the same as n increases. Margin of error increases as n increases. Margin of error decreases as n increases.

Solution

a)

Note that

E (margin of error) = z * s / sqrt(n)

Thus, as for a 95% confidence level,

z = 1.959963985

Then, if s = 7, n = 92,

E = 1.430382596 [answer]

b)

Using the same formula, if

n = 413, E = 0.675104659
n = 1598, E = 0.34320827 [answers]

c)

As you can see, MARGIN OF ERROR DECREASES AS N INCREASES. [OPTION C]