Answer :
Answer:
If we compare the p value obtained and the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of chips that do not fail in the first 1000 hours of their use is significantly higher than 0.57 or 57%.
Step-by-step explanation:
1) Data given and notation
n=1600 represent the random sample taken
X=0.6*1600=960 represent the number of chips do not fail in the first 1000 hours of their use
[tex]\hat p=\frac{960}{1600}=0.6[/tex] estimated proportion of chips do not fail in the first 1000 hours of their use
[tex]p_o=0.57[/tex] is the value that we want to test
[tex]\alpha=0.01[/tex] represent the significance level
Confidence=99% or 0.99
z would represent the statistic (variable of interest)
[tex]p_v[/tex] 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 above 57% of chips do not fail in the first 1000 hours of their use :
Null hypothesis: [tex]p\leq 0.57[/tex]
Alternative hypothesis:[tex]p > 0.57[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.6 -0.57}{\sqrt{\frac{0.57(1-0.57)}{1600}}}=2.42[/tex]
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 [tex]\alpha=0.01[/tex]. The next step would be calculate the p value for this test.
Since is a right tailed test the p value would be:
[tex]p_v =P(z>2.42)=0.0078[/tex]
If we compare the p value obtained and the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of chips that do not fail in the first 1000 hours of their use is significantly higher than 0.57 or 57%.