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Vlad1618 [11]
3 years ago
5

A student solved this problem and said the answer is 5 pounds. Todd has two packages of hamburger. One package weighs 4/5 pound

and the other weighs 9/10 pound. What is the total weight of both packages of hamburger combined? Is the student's answer reasonable? A. Yes, the answer is reasonable. B. No, the answer is not reasonable. It should be about 3 lb. C. No, the answer is not reasonable. It should be about 2 lb. D. No, the answer is not reasonable. It should be about 1 lb.
Mathematics
1 answer:
Lubov Fominskaja [6]3 years ago
5 0
The answer is C .This is because 4/5 is equal to 8/10. 9/10 plus 8/10 equals to 17/10 which is 1 7/10.Which is about 2 lb
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z=\frac{0.467 -0.9}{\sqrt{\frac{0.9(1-0.9)}{15}}}=-5.59  

p_v =P(z  

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 germinated seeds is significantly lower than 0.9 or 90%

Step-by-step explanation:

Data given and notation

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X=7 represent the number of seeds germinated

\hat p=\frac{7}{15}=0.467 estimated proportion of seeds germinated

p_o=0.9 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)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion of germinated seeds is less than 0.9 or 90%.:  

Null hypothesis:p\geq 0.9  

Alternative hypothesis:p < 0.9  

When we conduct a proportion test we need to use the z statisitc, 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.

Calculate the statistic  

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

z=\frac{0.467 -0.9}{\sqrt{\frac{0.9(1-0.9)}{15}}}=-5.59  

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

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p_v =P(z  

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