Question

Scientists think that robots will play a crucial role in factories in the next several decades. Suppose that in an experiment to determine whether the use of robots to weave computer cables is feasible, a robot was used to assemble 490 cables. The cables were examined and there were 11 defectives. If human assemblers have a defect rate of 0.035 (3.5%), does this data support the hypothesis that the proportion of defectives is lower for robots than humans

Answer 1

`z=\frac{0.0224 -0.035}{\sqrt{(0.035(1-0.035))/(490)}}=-1.518`

`p_v =P(z<-1.518)=0.0645`

If we compare the p value and the significance level assumed
`\alpha=0.05` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of defectives for robots is not significantly lower than the proportion of defectives for human

Explanation:

Data given and notation

n=490 represent the random sample taken

X=11 represent the number of defectives for robots

`\hat p=(11)/(490)=0.0224` estimated proportion of defectives for robots

`p_o=0.035` is the value that we want to test

`\alpha` represent the significance level

z would represent the statistic (variable of interest)

`p_v` represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion of defectives for robots is lower than for the humans :

Null hypothesis:
`p \geq 0.035`

Alternative hypothesis:
`p < 0.035`

When we conduct a proportion test we need to use the z statistic, and the is given by:

`z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}}` (1)

The One-Sample Proportion Test is used to assess whether a population proportion
`\hat p` is significantly different from a hypothesized value
`p_o`.

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

`z=\frac{0.0224 -0.035}{\sqrt{(0.035(1-0.035))/(490)}}=-1.518`

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level assumed for this case is
`\alpha=0.05`. The next step would be calculate the p value for this test.

Since is a lower tail test test the p value would be:

`p_v =P(z<-1.518)=0.0645`

If we compare the p value and the significance level assumed
`\alpha=0.05` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of defectives for robots is not significantly lower than the proportion of defectives for human