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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
Let x = hours of service.
Let y = total charges for x hours
The fixed charge is $20.
The charge per hour is $20.
Brock's charges for x hours is
y = 40 + 20x
This is a straight line with slope = 20 and y-intercept = 40.
The correct graph is W.
Answer: W
Let y = total charges for x hours
The fixed charge is $20.
The charge per hour is $20.
Brock's charges for x hours is
y = 40 + 20x
This is a straight line with slope = 20 and y-intercept = 40.
The correct graph is W.
Answer: W