Question

In a random sample of 500 television sets from a large production line, there were 80 defective sets. In a random sample of 200 television sets from a second production line, there were 10 defective sets. Answer questions A and B by following the sub-questions below. A Do these data provide sufficient evidence to establish that the proportion of defective sets from the first line exceeds the proportion of defective sets from the second line? (Use = 0.05.)

Answer 1

Answer:

$z=\frac{0.16-0.05}{\sqrt{0.129(1-0.129)(\frac{1}{500}+\frac{1}{200})}}=3.92$    

Now we can calculate the p value with the following probability:

$p_v =P(Z>3.92)= 0.000044$    

Since the p value is a very low value we have enough evidence to reject the null hypothesis and we can conclude that the true proportion of defectives from the firt line is significantly higher than the proportion of defective sets from the second line

Step-by-step explanation:

Information given

$X_{1}=80$ represent the number of defectives in the sample 1

$X_{2}=10$ represent the number of defectives in the sample 2

$n_{1}=500$ sample 1 selected  

$n_{2}=200$ sample 2 selected  

$p_{1}=\frac{80}{500}=0.16$ represent the proportion defectives in the sample 1

$p_{2}=\frac{10}{200}=0.05$ represent the proportion estimated of defectives in the sample 2

$\hat p$ represent the pooled estimate of p

z would represent the statistic    

$p_v$ represent the value

$\alpha=0.05$ significance level given  

Hypothesis to test

We want to verify if he proportion of defective sets from the first line exceeds the proportion of defective sets from the second line, the system of hypothesis would be:    

Null hypothesis:$p_{1} \leq p_{2}$    

Alternative hypothesis:$p_{1} > p_{2}$    

The statistic is given by:

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

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{80+10}{500+200}=0.129$  

Replacing the info given we got:

$z=\frac{0.16-0.05}{\sqrt{0.129(1-0.129)(\frac{1}{500}+\frac{1}{200})}}=3.92$    

Now we can calculate the p value with the following probability:

$p_v =P(Z>3.92)= 0.000044$    

Since the p value is a very low value we have enough evidence to reject the null hypothesis and we can conclude that the true proportion of defectives from the firt line is significantly higher than the proportion of defective sets from the second line