Question

A cell phone store has sold 150 phones of brand A and had 14 returned as defective. Additionally it has sold 125 phones of Brand B and had 15 returned as defective. Is there statistical evidence that Brand A has a smaller chance of being returned than Brand B? Use the 5-step method.

Answer 1

Answer:

$z=\frac{0.12-0.093}{\sqrt{0.105(1-0.105)(\frac{1}{150}+\frac{1}{125})}}=0.726$  

$p_v =P(Z>0.726)= 0.468$  

So the p value is a very high value and using any significance level for example $\alpha=0.05, 0,1,0.15$ 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 defective from Brand B  is NOT significantly higher than the proportion of of defective from Brand A  

Step-by-step explanation:

1) Data given and notation  

$X_{A}=14$ represent the number of returned from brand A

$X_{B}=15$ represent the number of returned from brand B

$n_{A}=150$ sample for Brand A selected

$n_{B}=125$ sample for Brand B selected

$p_{A}=\frac{14}{150}=0.093$ represent the proportion of returned from brand A

$p_{WCB}=\frac{15}{125}=0.12$ represent the proportion of returned from brand B

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 proportion of defective for Brand B is higher than for Brand A , the system of hypothesis would be:  

Null hypothesis:$p_{B} \leq p_{A}$  

Alternative hypothesis:$p_{B} > p_{A}$  

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

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

Where $\hat p=\frac{X_{B}+X_{A}}{n_{B}+n_{A}}=\frac{14+15}{150+125}=0.105$

3) Calculate the statistic

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

$z=\frac{0.12-0.093}{\sqrt{0.105(1-0.105)(\frac{1}{150}+\frac{1}{125})}}=0.726$  

4) Statistical decision

For this case we don't have a significance level provided $\alpha$, but we can calculate the p value for this test.  

Since is a one right tailed test the p value would be:  

$p_v =2*P(Z>0.726)= 0.468$  

5) Conclusion

So the p value is a very high value and using any significance level for example $\alpha=0.05, 0,1,0.15$ 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 defective from Brand B  is NOT significantly higher than the proportion of of defective from Brand A