Answer:
0.665
Step-by-step explanation:
Given: 100 people are split into two groups 70 and 30. I group is given cough syrup treatment but second group did not.
Prob for a person to be in the cough medication group = 0.70
Out of people who received medication, 34% did not have cough
Prob for a person to be in cough medication and did not have cough
=
Prob for a person to be not in cough medication and did not have cough
=
Probability for a person not to have cough
= P(M1C')+P(M2C')
where M1 = event of having medication and M2 = not having medication and C' not having cough
This is because M1 and M2 are mutually exclusive and exhaustive
SO P(C') = 0.397+0.2=0.597
Hence required prob =P(M1/C') = 
25% 3/10 1/3 and 37.5%
.25,.30,.33,.375
Answer:
(x+2)(x+6)
Step-by-step explanation:
x2 + 8x +12
x 2
x 6
Answer:
Step-by-step explanation:
1. Null hypothesis: u <= 0.784
Alternative hypothesis: u > 0.784
2. Find the test statistics: z using the one sample proportion test. First we have to find the standard deviation
Using the formula
sd = √[{P (1-P)}/n]
Where P = 0.84 and n = 750
sd =√[{0.84( 1- 0.84)/750]}
sd=√(0.84 (0.16) /750)
SD =√(0.1344/750)
sd = √0.0001792
sd = 0.013
Then using this we can find z
z = (p - P) / sd
z = (0.84-0.784) / 0.013
z =(0.056/0.013)
z = 4.3077
3. Find the p value and use it to make conclusions...
The p value at 0.02 level of significance for a one tailed test with 4.3077 as z score and using a p value calculator is 0.000008254.
4. Conclusions: the results is significant at 0.02 level of significance suck that we can conclude that its on-time arrival rate is now higher than 78.4%.
