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almond37 [142]
3 years ago
5

What would be the answer to 5/4 + 2/-9? How would you solve that? And if you could, could u leave the steps and then put the ans

wer to the equation in the comments so I can try to work it out

Mathematics
1 answer:
Gemiola [76]3 years ago
7 0

Answer: 37/36

Step-by-step explanation: find the lowest common multiples of 4 and -9 and the answer is -36

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Answer:

z=\frac{0.18 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=2.67  

Step-by-step explanation:

1) Data given and notation

n=100 represent the random sample taken

X=18 represent the ounce cups of coffee that were underfilled

\hat p=\frac{18}{100}=0.18 estimated proportion of ounce cups of coffee that were underfilled

p_o=0.1 is the value that we want to test

\alpha represent the significance level  

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 the true proportion it's higher than 0.1 or 10%:  

Null hypothesis:p\leq 0.1  

Alternative hypothesis:p > 0.1  

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{0.18 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=2.67  

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

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

p_v =P(Z>2.67)=0.0037  

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 thetrue proportion is not significanlty higher than 0.1 or 10% .  

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