Answer:
The remaining 15 athletes did not participate in any type of flexibility program
Answer:
Yes, result is significant ; PVALUE < α
Step-by-step explanation:
Given :
x = 536
n = sample size = 1012
Phat = x / n = 536 / 1012 = 0.5296 = 0.53
H0 : P0 = 0.5
H1 : P0 > 0.5
Test statistic :
(Phat - P0) ÷ sqrt[(P0(1 - P0)) / n]
1-P0 = 1 - 0.5 = 0.5
(0.53 - 0.5) ÷ sqrt[(0.5*0.5)/1012]
0.03 ÷ 0.0157173
= 1.9087
Pvalue :
Using the Pvalue from test statistic :
Pvalue = 0.02815
To test if result is significant :
α = 0.05
0.02815 < 0.05
Pvalue < α ; Hence, result is significant at α=0.05; Hence, we reject H0.
Answer: this is hard I hate math sorry
Step-by-step explanation:
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