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

Answer:

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

$p_v =P(z$  

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.

Explanation:

Data given and notation

n=100 represent the random sample taken

X=92 represent the successful

$\hat p=\frac{92}{100}=0.92$ 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:  

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (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:  

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

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:  

$p_v =P(z$  

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.