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

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Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ1 = μ2
Alternative Hypothesis, Ha: μ1 ≠ μ2

Rejection Region
This is two tailed test, for α = 0.05 and df = n1 + n2 - 2 = 98
Critical value of t are -1.984 and 1.984.
Hence reject H0 if t < -1.984 or t > 1.984

Pooled Variance
sp = sqrt((((n1 - 1)*s1^2 + (n2 - 1)*s2^2)/(n1 + n2 - 2))*(1/n1 + 1/n2))
sp = sqrt((((50 - 1)*37^2 + (50 - 1)*41^2)/(50 + 50 - 2))*(1/50 + 1/50))
sp = 7.8102

Test statistic,
t = (x1bar - x2bar)/sp
t = (209 - 225)/7.8102
t = -2.049

P-value Approach
P-value = 0.0431
As P-value >= 0.05, fail to reject null hypothesis.

b)

Given CI level is 0.95, hence α = 1 - 0.95 = 0.05                  
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 1.984

Pooled Variance
sp = sqrt((((n1 - 1)*s1^2 + (n2 - 1)*s2^2)/(n1 + n2 - 2))*(1/n1 + 1/n2))
sp = sqrt((((50 - 1)*37^2 + (50 - 1)*41^2)/(50 + 50 - 2))*(1/50 + 1/50))
sp = 7.8102                  
                  
Margin of Error                  
ME = tc * sp                  
ME = 1.984 * 7.8102                  
ME = 15.495                  
                  
CI = (x1bar - x2bar - tc * sp , x1bar - x2bar + tc * sp)                  
CI = (209 - 225 - 1.984 * 7.8102 , 209 - 225 - 1.984 * 7.8102                  
CI = (-31.4954 , -0.5046)