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Fantom [35]
3 years ago
10

Trials in an experiment with a polygraph include results that include cases of wrong results and cases of correct results. Use a

significance level to test the claim that such polygraph results are correct less than ​% of the time. Identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, P-value, conclusion about the null​ hypothesis, and final conclusion that addresses the original claim. Use the​ P-value method. Use the normal distribution as an approximation of the binomial distribution.
Mathematics
1 answer:
hram777 [196]3 years ago
5 0

Answer and Step-by-step explanation:

This is a complete question

Trials in an experiment with a polygraph include 97 results that include 23 cases of wrong results and 74 cases of correct results. Use a 0.01 significance level to test the claim that such polygraph results are correct less than 80​% of the time. Identify the null​hypothesis, alternative​ hypothesis, test​ statistic, P-value, conclusion about the null​ hypothesis, and final conclusion that addresses the original claim. Use the​ P-value method. Use the normal distribution as an approximation of the binomial distribution.

The computation is shown below:

The null and alternative hypothesis is

H_0 : p = 0.80

Ha : p < 0.80

\hat p = \frac{x}{ n} \\\\= \frac{74}{97}

= 0.7629

Now Test statistic = z

= \hat p - P0 / [\sqrtP0 \times (1 - P0 ) / n]

= 0.7629 - 0.80 / [\sqrt(0.80 \times 0.20) / 97]

= -0.91

Now

P-value = 0.1804

\alpha = 0.01

P-value > \alpha

So, it is Fail to reject the null hypothesis.

There is ample evidence to demonstrate that less than 80 percent of the time reports that these polygraph findings are accurate.

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Calculate 11010-1101 in base 2​
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Answer:

1101 base 2

Step-by-step explanation:

you first take the larger number up the the lower number down

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3 years ago
If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar
Oksana_A [137]

Answer:

n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}

Step-by-step explanation:

Lets divide it in cases, then sum everything

Case (1): All 5 numbers are different

 In this case, the problem is reduced to count the number of subsets of cardinality 5 from a set of cardinality n. The order doesnt matter because once we have two different sets, we can order them descendently, and we obtain two different 5-tuples in decreasing order.

The total cardinality of this case therefore is the Combinatorial number of n with 5, in other words, the total amount of possibilities to pick 5 elements from a set of n.

{n \choose 5 } = \frac{n!}{5!(n-5)!}

Case (2): 4 numbers are different

We start this case similarly to the previous one, we count how many subsets of 4 elements we can form from a set of n elements. The answer is the combinatorial number of n with 4 {n \choose 4} .

We still have to localize the other element, that forcibly, is one of the four chosen. Therefore, the total amount of possibilities for this case is multiplied by those 4 options.

The total cardinality of this case is 4 * {n \choose 4} .

Case (3): 3 numbers are different

As we did before, we pick 3 elements from a set of n. The amount of possibilities is {n \choose 3} .

Then, we need to define the other 2 numbers. They can be the same number, in which case we have 3 possibilities, or they can be 2 different ones, in which case we have {3 \choose 2 } = 3  possibilities. Therefore, we have a total of 6 possibilities to define the other 2 numbers. That multiplies by 6 the total of cases for this part, giving a total of 6 * {n \choose 3}

Case (4): 2 numbers are different

We pick 2 numbers from a set of n, with a total of {n \choose 2}  possibilities. We have 4 options to define the other 3 numbers, they can all three of them be equal to the biggest number, there can be 2 equal to the biggest number and 1 to the smallest one, there can be 1 equal to the biggest number and 2 to the smallest one, and they can all three of them be equal to the smallest number.

The total amount of possibilities for this case is

4 * {n \choose 2}

Case (5): All numbers are the same

This is easy, he have as many possibilities as numbers the set has. In other words, n

Conclussion

By summing over all 5 cases, the total amount of possibilities to form 5-tuples of integers from 1 through n is

n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}

I hope that works for you!

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Answer:

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