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svet-max [94.6K]
3 years ago
11

Rewrite 85% as a fraction in simplest form and as a decimal

Mathematics
1 answer:
nignag [31]3 years ago
3 0

85 %

85/100 - divide by 5 for the numerator and denominator

85/5= 17

100/5=20

Simplest form : 17/20

As decimal : 0.85 because of 85/100

Answer:

Simplest form : 17/20

As decimal : 0.85

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A sample of 1200 computer chips revealed that 45% of the chips fail in the first 1000 hours of their use. The company's promotio
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Answer:

z=\frac{0.45 -0.48}{\sqrt{\frac{0.48(1-0.48)}{1200}}}=-2.08

p_v = P(Z

So the p value obtained was a low value and using the significance level given \alpha=0.05 we see that 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 chips that fail in the first 1000 hours of their use is not significantly less than 0.48.   

Step-by-step explanation:

Data given and notation

n=1200 represent the random sample taken

\hat p=0.45 estimated proportion of chips that fail in the first 1000 hours of their use

\mu_0 =0.48 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)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion si less then 0.48:  

Null hypothesis:p\geq 0.48  

Alternative hypothesis:p < 0.48  

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  is significantly different from a hypothesized value .

Calculate the statistic  

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

z=\frac{0.45 -0.48}{\sqrt{\frac{0.48(1-0.48)}{1200}}}=-2.08

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 left tailed test the p value would be:  

p_v = P(Z

So the p value obtained was a low value and using the significance level given \alpha=0.05 we see that 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 chips that fail in the first 1000 hours of their use is not significantly less than 0.48.  

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4 years ago
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