Construct a 95% confidence interval for the mean difference of the population given the following information. Would you reject or fail to reject the claim that there is no difference in the mean?
$d̄=-0.728$
$s_{d}=1.34$
$n=10$

A personal trainer is studying whether a new stretching routine improves flexibility. She records the forward reach (in cm) of 6 clients before and after a 4-week program. Calculate the difference (after - before) for each client, the mean difference, and standard deviation.

A personal trainer is studying whether a new stretching routine improves flexibility. She records the forward reach (in cm) of 6 clients before and after a 4-week program. Calculate the difference (after - before) for each client, the mean difference, and standard deviation.