Answer: It looks like Part B is about 4 times that of part A.
Step-by-step explanation: I think the answer would be 12.
A would be 3. B would be 12.
I'm terribly sorry if it is wrong.
Answer:
z= 3.63
z for significance level = 0.05 is ± 1.645
Step-by-step explanation:
Here p = 42% = 0.42
n= 500
We formulate our null and alternative hypotheses as
H0: p= 0.42 against Ha : p> 0.42 One tailed test
From this we can find q which is equal to 1-p= 1-0.42 = 0.58
Taking p`= 0.5
Now using the z test
z= p`- p/ √p(1-p)/n
Putting the values
z= 0.5- 0.42/ √0.42*0.58/500
z= 0.5- 0.42/ 0.0220
z= 3.63
For one tailed test the value of z for significance level = 0.05 is ± 1.645
Since the calculated value does not fall in the critical region we reject our null hypothesis and accept the alternative hypothesis that more than 42% people owned cats.
Nathan has an infection and he needs to be treated with penicillin
But there’s a 75% probability of him being allergic to penicillin
And to test if the skin reacts to penicillin, the test is 98% accuracy
So even if he has the allergy the test is only 98% accurate of identifying the allergy
Therefore we are asked to find the probability of both these events happening
Event 1 and event 2 both should happen then. When the ‘and’ function is used in probabilities then the probabilities of both events happening should be multiplied
Therefore probability that Nathan has the allergy and test predicts it is
= 75% x 98% = 0.735
The answer is D. 0.735
D = rt
d/r = t divide both sides by r
t = d/r <<< the equation
hope this helped, God bless!
