Answer: YES.
the evidence supports this claim.
Step-by-step explanation:
We have that from the complete information,
H0: p1 = p2
Ha : p1 > p2
N1 = 654, P1 = 103/654
N2 = 615, P2 = 52/615
First we have to check the states of typicality that is if n1p1 and n1*(1-p1) and n2*p2 and n2*(1-p2) all are more prominent than equivalent to 5 or not
N1*p1 = 103
N1*(1-p1) = 551
N2*p2 = 52
N2*(1-p2) = 563
All the conditions are met so we can utilize standard typical z table to direct the test
Test measurements z = (P1-P2)/standard mistake
Standard blunder = √{p*(1-p)}*√{(1/n1)+(1/n2)}
P = pooled extent = [(p1*n1)+(p2*n2)]/[n1+n2]
After replacement
Test measurements z = 3.97
From z table, P(Z>3.97) = 0.000036
Thus, P-Value = 0.000036
As the got P-Value is not exactly the given noteworthiness 0.01
We reject the invalid speculation
Thus, we have enough proof to help the claim.that a higher extent of subjects in bunch 1 experienced drowsiness.
cheers I hope this helped !!