Question

Two different types of polishing solutions are being evaluated for possible use in a tumble-polish operation for manufacturing interocular lenses used in the human eye following cataract surgery. Three hundred lenses were tumble polished using the first polishing solution, and of this number, 253 had no polishing-induced defects. Another 300 lenses were tumble-polished using the second polishing solution, and 196 lenses were satisfactory upon completion.
Is there any reason to believe that the two polishing solutions differ? Use α = 0.05. What is the P-value for this test?

Answer 1

Answer:

$z=\frac{0.843-0.653}{\sqrt{0.748(1-0.748)(\frac{1}{300}+\frac{1}{300})}}=5.358$    

$p_v =2*P(Z>5.358) = 4.2x10^{-8}$  

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that we have singificantly differences between the two proportions.  

Step-by-step explanation:

Data given and notation  

$X_{1}=253$ represent the number with no defects in sample 1

$X_{2}=196$ represent the number with no defects in sample 1

$n_{1}=300$ sample 1

$n_{2}=300$ sample 2

$p_{1}=\frac{253}{300}=0.843$ represent the proportion of number with no defects in sample 1

$p_{2}=\frac{196}{300}=0.653$ represent the proportion of number with no defects in sample 2

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 in the the two proportions, the system of hypothesis would be:  

Null hypothesis:$p_{1} - p_2}=0$  

Alternative hypothesis:$p_{1} - p_{2} \neq 0$  

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 $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{253+196}{300+300}=0.748$  

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:  

$z=\frac{0.843-0.653}{\sqrt{0.748(1-0.748)(\frac{1}{300}+\frac{1}{300})}}=5.358$    

Statistical decision

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

$p_v =2*P(Z>5.358) = 4.2x10^{-8}$  

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that we have singificantly differences between the two proportions.