Question

The National Health and Nutrition Examination Survey (NHANES) assesses the health and nutritional status of adults and children in the United States using data from interviews and direct physical examinations. From 1999 to 2002, 712 men between the ages of 20 and 29 were examined. The mean body weight in this sample was 183.0 pounds. The margin of error associated with this estimate for 95% confidence was plus or minus 3 pounds. What is the probability (expressed as a percent) that the true population mean weight of men between the ages of 20 and 29 is between 180 and 186 pounds?

		90%
		95%
		6%
		None of the above
		It's impossible to tell from the information given.

Answer 1

given data,

level of significance, α = 0.05
margin of error = 3 pounds

margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
from standard normal table, two tailed value of |t α/2| with n-1 = 711 d.f is 1.963
margin of error = 1.963 * standard error
3 = 1.963 * standard error
standard error = 3/1.963 = 1.528
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size =712
standard error = sd/ sqrt(n)
1.528 = s.d/sqrt (712)
standard deviation = 1.528 *sqrt(712)
standard deviation = 40.772

the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 183
standard Deviation ( sd )= 40.772

the probability (expressed as a percent) that the true population mean weight of men
between the ages of 20 and 29 is between 180 and 186 pounds.To find P(a < = Z < = b) = F(b) - F(a)
P(X < 180) = (180-183)/40.772
= -3/40.772 = -0.0736
= P ( Z <-0.0736) From Standard Normal Table
= 0.4707
P(X < 186) = (186-183)/40.772
= 3/40.772 = 0.0736
= P ( Z <0.0736) From Standard Normal Table
= 0.5293
P(180 < X < 186) = 0.5293-0.4707 = 0.0587
p(180<X<186) = 5.87%
Approximately p(180<X<186) =6%
option:C