Answer to Question #299184 in Statistics and Probability for secret

Question

A new sneaker claims that it can make male athletes jump higher. A sample of 41 male athletes were asked to jump once with their own sneakers on and once wearing the new sneakers. The average jump increased by 1.28 inches with the new sneakers with a standard deviation of 1.02 inches. What is the margin of error of the 95% confidence interval for the average jump increase?

Answer 1

Let $\mu_1 =$ avg. jump with new sneakers

$\mu_2 =$ avg jump with own sneakers

$H_0 : \mu_1 - \mu_2 =0$ vs $H_1 : \mu_1 - \mu_2 >0$

The margin of error is given by,

$Z_{ \alpha } ({\sigma \over \sqrt {n}})$$=1.64({1.02 \over \sqrt {41}})=0.26125$

Therefore, the margin of error will be 0.26125