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

`z=\frac{0.12-0.093}{\sqrt{0.105(1-0.105)((1)/(150)+(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

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)=(14)/(150)=0.093` represent the proportion of returned from brand A

`p_(WCB)=(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)((1)/(n_(B))+(1)/(n_(A)))}}` (1)

Where
`\hat p=(X_(B)+X_(A))/(n_(B)+n_(A))=(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)((1)/(150)+(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