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Anika [276]
3 years ago
13

on saturday night hank had w basketball games he got to the gym 20 min before his first game each game lasted 1 1\2 hours there

was a 30 mins break i. between the two games hank went home 10 mins after his last game how long was hank at the gym.. A 20 mins B 245mins C 270mins D 275mins
Mathematics
1 answer:
Ne4ueva [31]3 years ago
6 0
It would be 20+90+90+30+10= 245
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mafiozo [28]

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

p_v =P(z>2.10)=0.018  

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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

\hat p=\frac{40}{120}=0.333 estimated proportion of strawberries damaged

p_o=0.25 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 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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27 and 36 have a Greatest Common Factor of 9 so...
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