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, ( p1 ), will be no higher than the percentage of defectives found in a competitor's line, ( p2 ). To test this statement, the retailer took a random sample of 230 of the salesman's cellular phones and 250 of the competitor's cellular phones. The retailer found that 27 of the salesman's cellular phones and 20 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=0.1 for the test. Step 1 of 6 : State the null and alternative hypotheses for the test.

Null hypothesis:    

Alternative hypothesis:  

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

Comparing the p value and the significance level given we see that so then we have enough evidence to reject the null hypothesis and then the proportion of defectives for the retailer is significantly higher than the proportion of defectives for the competitor at a 10% of significance level used.

Step-by-step explanation:

Data given

represent the number of defectives from the retailer

represent the number of defectives from the competitor

sample for the retailer

sample for the competitor

represent the proportion of defectives for the retailer

represent the proportion of defectives for the competitor

represent the pooled estimate of p

z would represent the statistic (variable of interest)    

represent the value for the test (variable of interest)  

significance level given  

System of hypothesis

We need to conduct a hypothesis in order to check if the percentage of defective cellular phones found among his products, ( p1 ), will be no higher than the percentage of defectives found in a competitor's line, ( p2 ), the system of hypothesis would be:    

Null hypothesis:    

Alternative hypothesis:    

The statistic is given by:

  (1)  

Where  

Calculate the statistic  

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

P value

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

Comparing the p value and the significance level given we see that so then we have enough evidence to reject the null hypothesis and then the proportion of defectives for the retailer is significantly higher than the proportion of defectives for the competitor at a 10% of significance level used.

eveidence i thing that

step-by-step explanation: