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Kay [80]
3 years ago
9

Normal probability plots indicate that the sample data come from normal populations. Are the requirements to use the​ one-way AN

OVA procedure​ satisfied?
Mathematics
1 answer:
Mrrafil [7]3 years ago
4 0

Answer:

No it is not satisfied

Step-by-step explanation:

The one-way ANOVA is used to measure whether the difference between the means of two independent groups are statistically significant.

For it to be satisfied, there has to be one independent variable which is categorical and one dependent variable. The dependent variable on its own has to be a continuous variable

The assumptions are:

1. Equal variance between population

2. Independence between observations

3. The random samples have to be gotten from a normal population.

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Whic of the following are valid names for the triangle below? Check all that apply.
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Answer:

A, C, E

Step-by-step explanation:

As long as all the 3 letters(D, K, and M) are included, it should be a valid name. These are the 3 points in the triangle afterall.

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Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar.
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Answer:

5 - x

Step-by-step explanation:

Given:

f(x)=25-x^2

g(x)=x+5

\begin{aligned}\left(\dfrac{f}{g}\right)(x) & = \dfrac{f(x)}{g(x)}\\\\ & = \dfrac{25-x^2}{x+5}\\\\& = \dfrac{(5-x)(5+x)}{(x+5)}\\\\& = \dfrac{(5-x)(x+5)}{(x+5)}\\\\& = 5-x\end{aligned}

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2 years ago
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Maria and her friends ate 5/8 of the 16 oranges that were at Maria's house. which is a reasonable answer for 5/8 x 16? please he
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Answer:

10

Step-by-step explanation:

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What is 0.6 as a percent
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Answer:

60

Step-by-step explanation:

when you multiply 100 by .6 it equals 60

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A sample of 1200 computer chips revealed that 45% of the chips fail in the first 1000 hours of their use. The company's promotio
yaroslaw [1]

Answer:

z=\frac{0.45 -0.48}{\sqrt{\frac{0.48(1-0.48)}{1200}}}=-2.08

p_v = P(Z

So the p value obtained was a low value and using the significance level given \alpha=0.05 we see that 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 chips that fail in the first 1000 hours of their use is not significantly less than 0.48.   

Step-by-step explanation:

Data given and notation

n=1200 represent the random sample taken

\hat p=0.45 estimated proportion of chips that fail in the first 1000 hours of their use

\mu_0 =0.48 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)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion si less then 0.48:  

Null hypothesis:p\geq 0.48  

Alternative hypothesis:p < 0.48  

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  is significantly different from a hypothesized value .

Calculate the statistic  

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

z=\frac{0.45 -0.48}{\sqrt{\frac{0.48(1-0.48)}{1200}}}=-2.08

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 left tailed test the p value would be:  

p_v = P(Z

So the p value obtained was a low value and using the significance level given \alpha=0.05 we see that 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 chips that fail in the first 1000 hours of their use is not significantly less than 0.48.  

6 0
3 years ago
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