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shepuryov [24]
3 years ago
13

7. Let A = {0, 1, {0,1}}. Give an example of a partition S of the power set P(A) such that SI = 3.

Mathematics
1 answer:
Natalka [10]3 years ago
5 0

Answer:

S=\Bigg\{~~\bigg\{\emptyset\bigg\},~~\bigg\{ \{0\},\{1\},\{\{0,1\}\} \bigg\},~~\bigg\{ \{0,1\},\{0,\{0,1\}\},\{1,\{0,1\}\},\{0,1,\{0,1\}\} \bigg\}~~ \Bigg\}

Step-by-step explanation:

The power set P(A) is the set made of ALL subsets of A. In this case it will be a set of 8 elements:

\mathcal{P}(A)=\Bigg\{~ \emptyset,~ \bigg\{0\bigg\},~ \bigg\{1\bigg\},~ \bigg\{\{0,1\}\bigg\},~ \bigg\{0,1\bigg\}, ~\bigg\{0,\{0,1\}\bigg\},~ \bigg\{1,\{0,1\}\bigg\}, ~\bigg\{0,1,\{0,1\}\bigg\}~ \Bigg\}

(Notice A has 3 elements on it, and P(A) is made of all subsets of it, we first listed those subsets of A with 0 elements (the empty set only), then we listed those with 1 element, then those with 2 elements, and finally those with 3 elements (which is the set A itself) )

A partition of P(A) is distributing all the elements of P(A) into nonempty disjoint sets. Think of it as simply distributing the elements of P(A) into different groups. In our case we want the partition to have 3 elements, so we want to distribute the 8 elements of P(A) into 3 groups. On the first group we just put the empty set, on the second group we put the elements of P(A) made of 1 single element, and on the third group we put the remaining 4 elements of P(A).

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Answer:

t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971    

p_v =2*P(t_{(1699)}>1.971)=0.049  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

Step-by-step explanation:

Data given and notation  

\bar X=669 represent the sample mean

s=732 represent the sample standard deviation

n=1700 sample size  

\mu_o =68 represent the value that we want to test

\alpha=0. represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is different from 634, the system of hypothesis would be:  

Null hypothesis:\mu = 634  

Alternative hypothesis:\mu \neq 634  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=1700-1=1699  

Since is a two sided test the p value would be:  

p_v =2*P(t_{(1699)}>1.971)=0.049  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

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