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Dimas [21]
3 years ago
12

Complete each of these tables. What do you notice?

Mathematics
2 answers:
Ber [7]3 years ago
8 0

Answer:

Answer no 1

Step1: 8+(-8)

Step2:8-8=0

Answer no 2

Step 1: 8-8=0 eight cancel from eight

Answer no 3

Step 1: 8+(-5)

Step2: 8-5=3

Answer no 4

Step1: 8-5=3

Answer no 5

Step 1: 8+(-12)

Step 2: 8-12 = -4

Answer no 6

Step 1: 8-12= -4

Rama09 [41]3 years ago
3 0

Answer:

0

0

3

3

-4

-4

Step-by-step explanation:

they have the same value for each couple

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Last year, a comprehensive report stated that 28% of businesses in the northeast of Ohio were considered highly profitable. This
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Answer:

z=\frac{0.38 -0.28}{\sqrt{\frac{0.28(1-0.28)}{50}}}=1.575  

p_v =2*P(z>1.575)=0.115  

So the p value obtained was a very high value and using the significance level given \alpha=0.01 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 businesses were highly profitable is not significantly different from 0.28 or 28%.

Step-by-step explanation:

Data given and notation

n=50 represent the random sample taken

X=19 represent the businesses were highly profitable

\hat p=\frac{19}{50}=0.38 estimated proportion of businesses were highly profitable

p_o=0.28 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 of businesses were highly profitable is different from 0.28 or no, the system of hypothesis is.:  

Null hypothesis:p=0.28  

Alternative hypothesis:p \neq 0.28  

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 required we can replace in formula (1) like this:  

z=\frac{0.38 -0.28}{\sqrt{\frac{0.28(1-0.28)}{50}}}=1.575  

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=0.01. 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>1.575)=0.115  

So the p value obtained was a very high value and using the significance level given \alpha=0.01 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 businesses were highly profitable is not significantly different from 0.28 or 28%.

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