Answer:
That result implies that t(s) is in the acceptance region for H₀ then we accept H₀, there is not statistics difference between the LL team and the national average
Step-by-step explanation:
The National average number of runs scored by a LL team is
μ = 5.7
Sample Information:
size sample n = 5
sample average x = 7.4
sample standard deviation s = 2.88
Is required to investigate if that sample average is statistically different from the National average
We will do a test with 95 % of confidence Interval that means
significance level α = 5 % or α = 0.05.
The sample size is 5 then even when we assume normal distribution the sample size indicates that we need to use t-student distribution. Furthermore, as the question is if the sample average is different from the national the test will be a two-tail test.
Then α = 0.05 α/2 = 0.025
df = n - 1 df = 5 - 1 df = 4
Then from t-student table we get t(c) = 2.132
Hypothesis test:
Null Hypothesis H₀ x = μ
Alternative Hypothesis Hₐ x ≠ μ
To calculate t (s)
t(s) = ( x - μ ) / s/√n
t(s) = ( 7.4 - 5.7 )* 2.24 / 2.88
t(s) = 1.7* 2.24 / 2.88
t(s) = 1.32
Comparing t(s) and t(c)
1.32 < 2.132
That result implies that t(s) is in the acceptance region for H₀ then we accept H₀, there is not statistics difference between the LL team and the national average
