Need help please anyone
1 answer:
You might be interested in
Answer:
P-value is less than 0.999995 .
Step-by-step explanation:
We are given that a driver's education course compared 1,500 students who had not taken the course with 1,850 students who had.
Null Hypothesis, :
{means students who took the driver's education course and those who didn't took have same chances to pass the written driver's exam the first time}
Alternate Hypothesis, :
{means students who took the driver's education course were more likely to pass the written driver's exam the first time}
The test statistics used here will be two sample Binomial statistics i.e.;
~ N(0,1)
Here, = No. of students passed the exam ÷ No.of Students that had taken the course
=
Similarly,
=
= 1,850
= 1,500
Test Statistics = = -27.38
P-value is given by, P(Z > -27.38) = 1 - P(Z > 27.38)
Now, in z table the highest critical value given is 4.4172 which corresponds to the probability value of 0.0005%. Since our test statistics is way higher than this so we can only say that the p-value for an appropriate hypothesis test is less than 0.999995 .
Answer:
a)
Operating room x 1 2 3 4
P(x) 0.15 0.25 0.4 0.2
b)
The graph of probability distribution is in attached image.
c)
All the probabilities lies in the range (0 to 1) and the probabilities add up to 1 so, the computed probability distribution is the valid probability distribution.
Step-by-step explanation:
Number of operating rooms Days
1 3
2 5
3 8
4 4
a)
Sum of frequencies=1+2+3+4=10
Operating rooms x frequency f Relative frequency
1 3 3/20=0.15
2 5 5/20=0.25
3 8 8/20=0.4
4 4 4/20=0.2
So, using the relative frequency approach, a discrete probability distribution for number of operating rooms in use on any given day is
Operating room x 1 2 3 4
P(x) 0.15 0.25 0.4 0.2
b)
The graph of probability distribution is in attached image.
c)
There are two conditions for a probability distribution to be valid
1. All probability must ranges from 0 to 1.
2. All probabilities must add up to 1.
We can see that all the probabilities lies in the range (0 to 1), so, condition 1 is satisfied.
For condition 2,
sum[p(x)]= 0.15+0.25+0.4+0.2=0.4+0.6=1.
As the probabilities add up to 1, so the condition 2 is also satisfied.
Thus, the computed probability distribution is the valid probability distribution.
