Question

Since many people have trouble programming their dvd​ players, an electronics company has developed what it hopes will be easier instructions. the goal is to have at least 9595​% of customers succeed at being able to program their dvd players. the company tests the new system on 100100 ​people, 9292 of whom were successful. is this strong evidence that the new system fails to meet the​ company's goal? a​ student's test of this hypothesis is provided. how many mistakes can you​ find?

Answer 1

`z=\frac{0.92-0.95}{\sqrt{(0.95(1-0.95))/(100)}}=-1.376`

`p_v =P(z<-1.376)=0.084`

So the p value obtained was a very low value and using the significance level given
`\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 interest is not significantly lower than 0.95, so then the program works since we fail to reject the null hypothesis.

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

X=92 represent the successful

`\hat p=(92)/(100)=0.92` estimated proportion of successes

`p_o=0.95` is the value that we want to test

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 is lower than 0.95 or no.:

Null hypothesis:
`p\geq 0.95`

Alternative hypothesis:
`p < 0.95`

When we conduct a proportion test we need to use the z statisitc, 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 required we can replace in formula (1) like this:

`z=\frac{0.92-0.95}{\sqrt{(0.95(1-0.95))/(100)}}=-1.376`

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
`\alpha=0.05`. The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

`p_v =P(z<-1.376)=0.084`

So the p value obtained was a very low value and using the significance level given
`\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 interest is not significantly lower than 0.95, so then the program works since we fail to reject the null hypothesis.