Question

In a random sample of males, it was found that 28 write with their left hands and 210 do not. In a random sample of females, it was found that 62 write with their land hands and 435 do not. use a 0.05 significance level to test the claim that the rate of left-handedness among males is less than that among females.a. What are the null and alternative hypotheses?b. What is the test statistic? (Round your answer to 4 decimal places)C. What is the P-Value (Round your answers to 4 decimal places)D. Based on your results, does it appear as if the rate of left-handedness among males is less than that among females? why or why not?

Answer 1

Answer:

a) Null hypothesis:$p_{M} \geq p_{W}$  

Alternative hypothesis:$p_{M} < p_{W}$  

b) $z=\frac{0.118-0.125}{\sqrt{0.122(1-0.122)(\frac{1}{238}+\frac{1}{497})}}=-0.271$    

c)$p_v =P(Z$   

d) If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the proportion of men that write with their left hand is NOT significant lower than the proportion of female that write with their left hand

Step-by-step explanation:

1) Data given and notation  

$X_{M}=28$ represent the number of men that write with their left hand

$X_{W}=62$ represent the number of women  that write with their left hand

$n_{M}=210+28=238$ sample of male selected

$n_{W}=62+435=497$ sample of female selected

$p_{M}=\frac{28}{238}=0.118$ represent the proportion of men  that write with their left hand

$p_{W}=\frac{62}{497}=0.125$ represent the proportion of women that write with their left hand

$\alpha=0.05$ represent the significance level

z would represent the statistic (variable of interest)  

$p_v$ represent the value for the test (variable of interest)

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the rate of left-handedness among males is less than that among females , the system of hypothesis would be:  

Null hypothesis:$p_{M} \geq p_{W}$  

Alternative hypothesis:$p_{M} < p_{W}$  

We need to apply a z test to compare proportions, and the statistic is given by:  

$z=\frac{p_{M}-p_{W}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{M}}+\frac{1}{n_{W}})}}$   (1)

Where $\hat p=\frac{X_{M}+X_{W}}{n_{M}+n_{W}}=\frac{28+62}{238+497}=0.122$

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:  

$z=\frac{0.118-0.125}{\sqrt{0.122(1-0.122)(\frac{1}{238}+\frac{1}{497})}}=-0.271$  

4) Statistical decision

We have a significance level provided $\alpha=0.05$, and now we can calculate the p value for this test.  

Since is a one left side test the p value would be:  

$p_v =P(Z$  

If we compare the p value and the significance level given $\alpha=0.05, 0,1,0.15$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the proportion of men that write with their left hand is NOT significant loer than the proportion of female that write with their left hand .