Question

A salesman for a new manufacturer of cellular phones claims not only that they cost the retailer less but also that the percentage of defective cellular phones found among his products, ( p1p1 ), will be no higher than the percentage of defectives found in a competitor's line, ( p2p2 ). To test this statement, the retailer took a random sample of 135135 of the salesman's cellular phones and 105105 of the competitor's cellular phones. The retailer found that 1515 of the salesman's cellular phones and 55 of the competitor's cellular phones were defective. Does the retailer have enough evidence to reject the salesman's claim? Use a significance level of α=0

Answer 1

Answer:

$z=\frac{0.111-0.524}{\sqrt{0.292(1-0.292)(\frac{1}{135}+\frac{1}{105})}}=-6.985$    

$p_v =P(Z$  

Comparing the p value with the significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the proportion of defective for the manufacturer is lower than the proportion of defective for the competitor at 5% of significance

Step-by-step explanation:

Data given and notation  

$X_{1}=15$ represent the number of defective for the manufacturer

$X_{2}=55$ represent the number of defective for the competitor

$n_{1}=135$ sample for the manufacturer

$n_{2}=105$ sample for the competitor

$p_{PC}=\frac{15}{135}=0.111$ represent the proportion of defective for the manufacturer

$p_{2}=\frac{55}{105}=0.524$ represent the proportion of defective for the competitor

z would represent the statistic (variable of interest)  

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

$\alpha=0.05$ significance level given

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if p1<p2, the system of hypothesis would be:  

Null hypothesis:$p_{1} - p_{2} \geq 0$  

Alternative hypothesis:$p_{1} - p_{2} < 0$  

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

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$   (1)  

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{15+55}{135+105}=0.292$  

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

Calculate the statistic  

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

$z=\frac{0.111-0.524}{\sqrt{0.292(1-0.292)(\frac{1}{135}+\frac{1}{105})}}=-6.985$    

Statistical decision

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

$p_v =P(Z$  

Comparing the p value with the significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the proportion of defective for the manufacturer is lower than the proportion of defective for the competitor at 5% of significance