A doctor is measuring the average height of male students at a large college. The doctor measures the heights, in inches, of a sample of 40 male students from the baseball team. Using this data, the doctor calculates the 95% confidence interval (63.5, 74.4). Which one of the following conclusions is valid?

a. No conclusion can be drawn.
b. The doctor can be 95% confident that the mean height of male students at the college is between 63.5 inches and 74.4 inches.
c. 95% of the male students from the baseball team have heights between 63.5 inches and 74.4 inches.

For this case the  95% confidence interval is given (63.5 , 74.4) and we want to conclude about the result. For this case we can say that the true mean of heights for male students would be between 63.5 and 74.4. And the best answer would be:

b. The doctor can be 95% confident that the mean height of male students at the college is between 63.5 inches and 74.4 inches.

Step-by-step explanation:

Notation

represent the sample mean for the sample  

population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

  (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

(2)  

(3)  

In order to calculate the critical value we need to find first the degrees of freedom, given by:

For this case the  95% confidence interval is given (63.5 , 74.4) and we want to conclude about the result. For this case we can say that the true mean of heights for male students would be between 63.5 and 74.4. And the best answer would be:

b. The doctor can be 95% confident that the mean height of male students at the college is between 63.5 inches and 74.4 inches.

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

c. 1,233 miles