Question

A sports team compared two versions of a basketball (denoted as J and K) with respect to the time it takes to reach the hoop when thrown. They randomly selected 100 basketball players, and then they randomly assigned 50 to each basketball version. Basketball J has a sample mean of 209 and an SD of 37, while Basketball K has a sample mean of 225 and an SD of 41. It is known that the population variances are equal.

The sports team wants to know if there’s a significant difference between the two basketball versions with regards to the mean time they take to reach the hoop. To answer this, compute a 95% confidence interval for the difference in the mean time for the two basketball versions (J and K), and state hypothesis and conclusion

*Please don't copy other experts' solution

Answer 1

Pooled standard deviation Sp =

= 38.3705

95% Confidence interval is :-

( X̅1 - X̅2 ) ± t( α/2 , n1+n2-2) SP √( (1/n1) + (1/n2))

df = n1 + n2 - 2 = 100 + 50 - 2 = 148

find t critical value at 0.05 significance level with 148 df

t(α/2, n1 + n1 - 2) = t( 0.05/2, 100 + 50 - 2) = 1.976 (from T table)

( 209 - 225 ) ± 1.976 * 38.3705 √ ( (1/100) + (1/50))

Lower Limit = ( 209 - 225 ) - t(0.05/2 , 100 + 50 -2) 38.3705 √( (1/100) + (1/50))

Lower Limit = -29.132

Upper Limit = ( 209 - 225 ) + t(0.05/2 , 100 + 50 -2) 38.3705 √( (1/100) + (1/50))

Upper Limit = -2.868

95% Confidence Interval is ( -29.132 , -2.868 )