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, ( 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.

Answer 1

Answer:20%

Explanation:

Answer 2

Null hypothesis:
`p_(1) \leq p_(2)`

Alternative hypothesis:
`p_(1)> p_(2)`

`z=\frac{0.117-0.08}{\sqrt{0.0979(1-0.0979)((1)/(230)+(1)/(250))}}=1.363`

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

`p_v =P(Z>1.363)= 0.0864`

Comparing the p value and the significance level given we see that
`p_v <\alpha=0.1` 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.

Explanation:

Data given

`X_(1)=27` represent the number of defectives from the retailer

`X_(2)=20` represent the number of defectives from the competitor

`n_(1)=230` sample for the retailer

`n_(2)=3377` sample for the competitor

`p_(1)=(27)/(230)=0.117` represent the proportion of defectives for the retailer

`p_(2)=(20)/(250)=0.08` represent the proportion of defectives for the competitor

`\hat p` represent the pooled estimate of p

z would represent the statistic (variable of interest)

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

`\alpha=0.1` 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:
`p_(1) \leq p_(2)`

Alternative hypothesis:
`p_(1)> p_(2)`

The statistic is given by:

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

Where
`\hat p=(X_(1)+X_(2))/(n_(1)+n_(2))=(27+20)/(230+250)=0.0979`

Calculate the statistic

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

`z=\frac{0.117-0.08}{\sqrt{0.0979(1-0.0979)((1)/(230)+(1)/(250))}}=1.363`

P value

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

`p_v =P(Z>1.363)= 0.0864`

Comparing the p value and the significance level given we see that
`p_v <\alpha=0.1` 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.