The population standard deviation for the height of college football players is 2.7 inches. If we want to estimate a 95% confidence interval for the population mean height of these players with a 0.65 margin of error, how many randomly selected players must be surveyed? (Round up your answer to nearest whole number, do not include any decimals)

n = 66 (to the nearest whole number)

66 randomly selected players must be surveyed

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

x+/-M.E

M.E = zr/√n

Making n the subject of formula;

n = (zr/M.E)^2 1

Given that;

Mean = x

Standard deviation r = 2.7 inches

Number of samples = n

Confidence interval = 95%

z(at 95% confidence) = 1.96

Margin of error M.E = 0.65 inches

Substituting the given values into equation 1;

n = (zr/M.E)^2

n = (1.96×2.7/0.65)^2

n = 66.28464852071

n = 66 (to the nearest whole number)

66 randomly selected players must be surveyed

can you rephrase the question ? it is unclear : -)

step-by-step explanation: