Question

Various studies indicate that approximately 9% of the world’s population is left handed. You think this number is actually higher. You take an SRS of 221 people and find that 22 of them are left handed. Test your claim at the 5% significance level. Part a: State your null and alternative hypotheses. Part b: Sketch the rejection region. Part c: Calculate the test statistic. Plot this value in your sketch in part b. Part d: Determine the P-value for your test. Part e: State your conclusions clearly in complete sentences.

Answer 1

Answer:

a) Null hypothesis:$p=0.09$  

Alternative hypothesis:$p \neq 0.09$  

b) $z_{crit} =\pm 1.96$

And we can see on the figure attached that the rejection zone is in red

c) $z=\frac{0.0995 -0.09}{\sqrt{\frac{0.09(1-0.09)}{221}}}=0.493$  

d) $p_v =2*P(z>0.493)=0.622$  

e) Since the p value is higher than the significance level we have don't have enough evidence to conclude that the true proportion of people left handed is different from 9%

Step-by-step explanation:

Information given

n=221 represent the sample selected

X=22 represent the number of people that are left handed

$\hat p=\frac{22}{221}=0.0995$ estimated proportion of people left handed

$p_o=0.09$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic

$p_v$ represent the p value

Part a

We need to conduct a hypothesis in order to test the claim that the true proportion of people left handed is 0.09.:  

Null hypothesis:$p=0.09$  

Alternative hypothesis:$p \neq 0.09$  

Part b

For this case we need to find a critical value who accumulates on each tail $\alpha/2=0.025$ of the area and we got:

$z_{crit} =\pm 1.96$

And we can see on the figure attached that the rejection zone is in red

Part c

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

$z=\frac{0.0995 -0.09}{\sqrt{\frac{0.09(1-0.09)}{221}}}=0.493$  

On the second figure we have the calculated value in green

Part d

Since is a bilateral test the p value is:  

$p_v =2*P(z>0.493)=0.622$  

Part e

Since the p value is higher than the significance level we have don't have enough evidence to conclude that the true proportion of people left handed is not different from 9%