Question

Two different types of injection-molding machines are used to form plastic parts. Two random samples, each of size 300, are selected. 15 defective parts are found in the sample from machine 1 and 8 defective parts are found in the sample from machine 2. Is it reasonable to assume that both machines have the same defective rate? Use α = 0.05.

Answer 1

`z=\frac{0.05-0.027}{\sqrt{0.038(1-0.038)((1)/(300)+(1)/(300))}}=1.473`

`p_v =2*P(Z>1.473)= 0.141`

Comparing the p value with the significance level given
`\alpha=0.05` 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't say that the defective rate analyzed is significantly different between the two groups.

Explanation:

Data given and notation

`X_(1)=15` represent the number of defectives from machine 1

`X_(2)=8` represent the number of defectives from machine 2

`n_(1)=300` sample 1 selected

`n_(2)=300` sample 2 selected

`p_(1)=(15)/(300)=0.05` represent the proportion of defectives for machine 1

`p_(2)=(8)/(300)=0.027` represent the proportion of defectives for machine 2

`\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=0.05` significance level given

Concepts and formulas to use

We need to conduct a hypothesis in order to check if is there is a difference between the two proportions of defective rates, the system of hypothesis would be:

Null hypothesis:
`p_(1) = p_(2)`

Alternative hypothesis:
`p_(1) \\eq 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))=(15+8)/(300+300)=0.038`

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{0.05-0.027}{\sqrt{0.038(1-0.038)((1)/(300)+(1)/(300))}}=1.473`

Statistical decision

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

`p_v =2*P(Z>1.473)= 0.141`

Comparing the p value with the significance level given
`\alpha=0.05` 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't say that the defective rate analyzed is significantly different between the two groups.