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LekaFEV [45]
3 years ago
11

Will mark brainliest Question is in attachment

Mathematics
1 answer:
ahrayia [7]3 years ago
4 0
Answer: angle QPR = 25
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Hi, all we need to do is to divide the number of brownies by the number of students, and we will get the amount of how many would each student get.

26/15 = 1,73 brownies per student.

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Step-by-step explanation:

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Read 2 more answers
Is this less than <br> Greater than<br> Or equal too?
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The values are equal.

Root 5.76= 2.4

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4 0
2 years ago
The number of events is 29​, the number of trials is 298​, the claimed population proportion is​ 0.10, and the significance leve
Nina [5.8K]

Answer:

z=\frac{0.0973 -0.1}{\sqrt{\frac{0.1(1-0.1)}{298}}}=-0.155  

p_v =2*P(Z  

And we can use excel to find the p value like this: "=2*NORM.DIST(-0.155;0;1;TRUE)"

So the p value obtained was a very high value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is not significantly different from 0.1 .  

Step-by-step explanation:

1) Data given and notation

n=298 represent the random sample taken

X=29 represent the events claimed

\hat p=\frac{29}{298}=0.0973 estimated proportion

p_o=0.1 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

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 proportion is 0.1 or no.:  

Null hypothesis:p=0.1  

Alternative hypothesis:p \neq 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{0.0973 -0.1}{\sqrt{\frac{0.1(1-0.1)}{298}}}=-0.155  

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

Since is a bilateral test the p value would be:  

p_v =2*P(Z  

And we can use excel to find the p value like this: "=2*NORM.DIST(-0.155;0;1;TRUE)"

So the p value obtained was a very high value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is not significantly different from 0.1 .  

We can do the test also in R with the following code:

> prop.test(29,298,p=0.1,alternative = c("two.sided"),conf.level = 1-0.05,correct = FALSE)

7 0
2 years ago
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