Question

In a survey of 100 randomly selected professional soccer players, the mean counter movement jump height was x=55.08cm. Suppose that somehow we knew that the population standard deviation of the counter movement jump is 5.52 cm.

Find a 90% confidence interval estimate of the mean CMJ of all professional soccer players.

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 55.08 cm

Population standard deviation = $\sigma$ = 5.52 cm

Sample size = n = 100

At 90% confidence level the z is ,

$\alpha$ = 1 - 90% = 1 - 0.90 = 0.1

$\alpha$ / 2 = 0.1 / 2 = 0.05

Z_$\alpha$/2 = Z_0.05 = 1.645

Margin of error = E = Z_$\alpha$/2* ($\sigma$ /$\sqrt$n)

= 1.645 * (5.52 / $\sqrt$ 100)

= 0.91

At 90% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

55.08 - 0.91 < $\mu$ < 55.08 + 0.91

54.17 < $\mu$ < 55.99

(54.17 , 55.99)