Various studies indicate that approximately 11% of the world's population is left handed. You think this number is actually higher. You take an SRS of 225 people and find that 31 of them are left handed. Test your claim at the 5% significance level. A. State your null and alternative hypotheses.
B. Sketch the rejection region.
C. Calculate the test statistic.
D. Determine the P-value for your test.

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

hello i am doing good for you that it's summer in your country or state ; )

btw in my country we just have sun and rain as (because i live in the caribbean lol)

also, thx a lot for the points : )