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Morgarella [4.7K]
2 years ago
6

Greenfields is a mail order seed and plant business. The size of orders is uniformly distributed over the interval from $25 to $

80. Use the following random numbers to generate the size of 10 orders. .41 .99 .07 .05 .38 .77 .19 .12 .58 .60 What is the value for the first order size generated randomly based on random number 0.41? What is the value for the last order size generated randomly based on random number 0.60? What is the average order size?
Mathematics
1 answer:
LekaFEV [45]2 years ago
4 0

Answer:

a) 47.55

b) 58

c) 47.88

Step-by-step explanation:

Given that the size of the orders is uniformly distributed over the interval

$25 ( a ) to $80 ( b )

<u>a) Determine the value for the first order size generated based on 0.41</u>

parameter for normal distribution is given as ;  a = 25,  b = 80

size/value of order  = a + random number ( b - a )

                                = 25 + 0.41 ( 80 - 25 )

                                =  47.55

<u>b) Value of the last order generated based on random number (0.6)</u>

= a + random number ( b - a )

= 25 + 0.6 ( 80 - 25 )

= 25 + 33 = 58

<u>c) Average order size </u>

= ∑ order 1 + order 2 + ----- + order 10  ) / 10

= (47.55 + ...... + 58 ) / 10

= 478.8 / 10 = 47.88

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Answer:

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Step-by-step explanation:

Data given and notation

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X=67 represent the applicants who request financial aid

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p_o=0.35 is the value that we want to test

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Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion of applicants who request financial aid is higher than 0.35.:  

Null hypothesis:p\leq 0.35  

Alternative hypothesis:p > 0.35  

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{0.447 -0.35}{\sqrt{\frac{0.35(1-0.35)}{150}}}=2.491  

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>2.491)=0.0064  

So the p value obtained was a very low value and using the significance level assumed \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 applicants who request financial aid is significantly higher than 0.35

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Attached

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Inequalities graph : brainly.com/question/11234618

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