Answer:
a. H0 : p ≤ 0.11 Ha : p >0.11 ( one tailed test )
d. z= 1.3322
Step-by-step explanation:
We formulate our hypothesis as
a. H0 : p ≤ 0.11 Ha : p >0.11 ( one tailed test )
According to the given conditions
p`= 31/225= 0.1378
np`= 225 > 5
n q` = n (1-p`) = 225 ( 1- 31/225)= 193.995> 5
p = 0.4 x= 31 and n 225
c. Using the test statistic
z= p`- p / √pq/n
d. Putting the values
z= 0.1378- 0.11/ √0.11*0.89/225
z= 0.1378- 0.11/ √0.0979/225
z= 0.1378- 0.11/ 0.02085
z= 1.3322
at 5% significance level the z- value is ± 1.645 for one tailed test
The calculated value falls in the critical region so we reject our null hypothesis H0 : p ≤ 0.11 and accept Ha : p >0.11 and conclude that the data indicates that the 11% of the world's population is left-handed.
The rejection region is attached.
The P- value is calculated by finding the corresponding value of the probability of z from the z - table and subtracting it from 1.
which appears to be 0.95 and subtracting from 1 gives 0.04998
Answer:
the answer is 62!
Step-by-step explanation:
multiply
Answer:
Step-by-step explanation:
Answer:
285
Step-by-step explanation:
hi
personally this is how i would do this question but there are other ways.
i would start off by seeing how many sundays there are in the month (which is 5) then the number of other days which is 25. i'd then multiply 5 by 510 which is 2550. add that to 240 times 25 (which is 6000) and all of that together is 8550. divide that by the number of the days in the month (30) and then you'll get the daily avergae for the month
hope this helped:)
If it was one year it would be 2200 and you just keep adding 200 for each year.