Question

Suppose National Collegiate Athletic Association [NCAA] rules state all student-athletes are to receive an average of 50 hours of academic support, per term. A random sample of 49 University of Maryland student-athletes revealed a mean of 47.5 hours of academic support per term. If the calculated value for the associated test statistic equaled -1.75, what was the standard deviation of the number of hours of academic support the student-athletes in the sample received per term?

Answer 1

Solution

Given that,

mean = $\mu$ = 50

$\bar x$ = 47.5

n = 49

z = -1.75

Using z-score formula  

$\bar x$ = z * $\sigma$ $\bar x$ + $\mu$ $\bar x$  

47.5 = -1.75 *$\sigma$$\bar x$ + 50

$\sigma$$\bar x$  = 47.5 - 50 / -1.75

$\sigma$$\bar x$  = 1.43

standard deviation = $\sigma$ = $\sigma$ $\bar x$ * $\sqrt{}$ n

standard deviation = $\sigma$ = 1.43 * $\sqrt{}$ 49

standard deviation = $\sigma$ = 10