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 180180 of the salesman's cellular phones and 225225 of the competitor's cellular phones. The retailer found that 1717 of the salesman's cellular phones and 99 of the competitor's cellular phones were defective. Does the retailer have enough evidence to reject the salesman's claim? Use a significanc

Answer 1

`z=\frac{0.094-0.04}{\sqrt{0.0642(1-0.0642)((1)/(180)+(1)/(225))}}=2.203`

`p_v =P(Z>2.203)= 0.0138`

Comparing the p value with the significance level given
`\alpha=0.01` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so then the claim from the retailer makes sense at 1% of significance.

Explanation:

Use a significance level of α=0.01 for the test.

Data given and notation

`X_(1)=17` represent the number of defectives from the retailer

`X_(2)=9` represent the number of defectives from the competitor

`n_(1)=180` sample 1 selected

`n_(2)=225` sample 2 selected

`p_(1)=(17)/(180)=0.094` represent the proportion estimated for defectives from the retailer

`p_(2)=(9)/(225)=0.04` represent the proportion estimated for defectives from 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=0.01` significance level given

Concepts and formulas to use

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)`

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)((1)/(n_(1))+(1)/(n_(2)))}}` (1)

Where
`\hat p=(X_(1)+X_(2))/(n_(1)+n_(2))=(17+9)/(180+225)=0.0642`

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{0.094-0.04}{\sqrt{0.0642(1-0.0642)((1)/(180)+(1)/(225))}}=2.203`

Statistical decision

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

`p_v =P(Z>2.203)= 0.0138`

Comparing the p value with the significance level given
`\alpha=0.01` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so then the claim from the retailer makes sense at 1% of significance.