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KatRina [158]
3 years ago
13

PLEASE HELP!! I WILL GIVE EXTRA POINTS!

Mathematics
1 answer:
inysia [295]3 years ago
7 0
1. B, dependent variable

2. C, the dependent variable is represented in the 2nd column of a table.

8.
A. Domain: { 0, -3, 4, 5 }
B. Range: { 0, 1, 6, 7, 8 }
C. No, this is not a function because the x-value of 4 (domain) has two corresponding y-values of 6 and 8 (range)
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A buyer went to the market to buy strawberries. He purchased 120 randomly selected strawberries from a vendor who claimed that n
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Answer:

z=\frac{0.333 -0.25}{\sqrt{\frac{0.25(1-0.25)}{120}}}=2.10  

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Step-by-step explanation:

Data given and notation

n=120 represent the random sample taken

X=40 represent the number of strawberries damaged

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Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that no more than 25% of his total harvest of strawberries was damaged.:  

Null hypothesis:p\leq 0.25  

Alternative hypothesis:p > 0.25  

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.

Calculate the statistic  

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

z=\frac{0.333 -0.25}{\sqrt{\frac{0.25(1-0.25)}{120}}}=2.10  

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 next step would be calculate the p value for this test.  

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

p_v =P(z>2.10)=0.018  

At 5% of significance we can conclude that the true proportion of strawberries damage is higher than 0.25

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