Mathematics, 30.11.2019 01:31 jacob7542

Asample of 1600 computer chips revealed that 54% of the chips do not fail in the first 1000 hours of their use. the company's promotional literature claimed that more than 51% do not fail in the first 1000 hours of their use. is there sufficient evidence at the 0.05 level to support the company's claim?

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Answer from: enm01

p_v =P(z2.40)=0.0082

The p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have p_v 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

$\hat p=0.54$ estimated proportion of the chips do not fail in the first 1000 hours of their use

p_o=0.51 is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v 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: $p\leq 0.51$

Alternative hypothesis: p 0.51

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

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

3) Calculate the statistic

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

$z=\frac{0.54 -0.51}{\sqrt{\frac{0.51(1-0.51)}{1600}}}=2.40$

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

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

p_v =P(z2.40)=0.0082

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have p_v 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%.

Answer from: Quest

8 2/3 is the answer

step-by-step explanation: just use a calculator

Answer from: Quest

roots are from trees

step-by-step explanation:

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