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n200080 [17]
3 years ago
12

I am interested in determining how long students take on average, to finish the first exam in my class. I specifically am intere

sted in looking at students who take the class during the summer semester. I randomly pick 50 students from previous summer semesters and note how long the took on the first exam (TIME), which class they took with me - Stator Stat (CLASS), and whether they took the class online or in person (FORMAT).
Identity the following viables as either quantitative or qualitative
Class
Time
Mathematics
1 answer:
Advocard [28]3 years ago
6 0

Answer:

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Step-by-step explanation:

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Justin wants his new treehouse to blend in with the leaves, so he decides to paint it green. He mixes some white paint and green
spayn [35]

Answer:

  • C. neither. the roof and walls are the same shade of green

Step-by-step explanation:

<u>The first batch:</u>

  • Some white and some green

<u>The second batch:</u>

  • Triple amount of white and triple amount of green

As we see the ratio of white to green has not changed

Therefore there won't be any difference in painting

Correct answer choice is C.

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Order of operations (19-8)×(10-5)+8
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PEMDAS
(9)(5) + 8
45 + 8
53
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What statement makes the open sentence 2.4 = 6x true?
Softa [21]
C. 0.4 times 6 is 2.4. Hope that helps.
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What is the value of
Len [333]

n-2=10n+42\qquad\text{add 2 to both sides}\\\\n=10n+44\qquad\text{subtract 10n from both sides}\\\\-9n=44\qquad\text{divide both sides by (-9)}\\\\\boxed{n=-\dfrac{44}{9}}

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3 years ago
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4.One attorney claims that more than 25% of all the lawyers in Boston advertise for their business. A sample of 200 lawyers in B
AleksAgata [21]

Answer:

z=\frac{0.315 -0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}}=2.123  

p_v =P(Z>2.123)=0.0169  

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 lawyers had used some form of advertising for their business is significantly higher than 0.25 or 25% .  

Step-by-step explanation:

1) Data given and notation  

n=200 represent the random sample taken

X=63 represent the lawyers had used some form of advertising for their business

\hat p=\frac{63}{200}=0.315 estimated proportion of lawyers had used some form of advertising for their business

p_o=0.25 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 more than 25% of all the lawyers in Boston advertise for their business:  

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.

3) Calculate the statistic  

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

z=\frac{0.315 -0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}}=2.123  

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 significance level provided \alpha=0.05. 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>2.123)=0.0169  

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 lawyers had used some form of advertising for their business is significantly higher than 0.25 or 25% .  

8 0
3 years ago
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