A student will spend 15 hours average each week in class. If the standard deviation is 2 hours, calculate the z-score for a student who spends less than 12 hours in class?

The p value obtained was a very low value and using the significance level given we have so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of the chips that do not fail in the first 1000 hours of their use is higher than 0.51 or 51% (Company's claim).  

Step-by-step explanation:

1) Data given and notation n  

n=1600 represent the random sample taken

X=represent the number of the chips do not fail in the first 1000 hours of their use

estimated proportion of the chips do not fail in the first 1000 hours of their use

is the value that we want to test

represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that more than 51% do not fail in the first 1000 hours of their use, the system of hypothesis on this case would be:

Null hypothesis:  

Alternative hypothesis:  

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

(1)  

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

3) Calculate the statistic  

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

4) 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 provided . The next step would be calculate the p value for this test.  

Since is a upper right tailed test the p value would be:  

So the p value obtained was a very low value and using the significance level given we have so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of the chips that do not fail in the first 1000 hours of their use is higher than 0.51 or 51%.