Question

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.

Answer 1

a. H0 : p ≤ 0.11 Ha : p >0.11 ( one tailed test )

d. z= 1.3322

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