In randomized, double-blind clinical trials of a new vaccine, were randomly divided into two groups. Subjects in group 1 received the new vaccine while subjects in group 2 received a control vaccine. After the second dose, of subjects in the experimental group (group 1) experienced as a side effect. After the second dose, of of the subjects in the control group (group 2) experienced as a side effect. Does the evidence suggest that a higher proportion of subjects in group 1 experienced as a side effect than subjects in group 2 at the level of significance?

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 !!

in simplified version   it would be x ≤-4

step-by-step explanation:

multiply each term in x ≤-4 by -1

the final price would be 29.6925

step-by-step explanation:

39.59 minus 25% equals 29.6925