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Gemiola [76]
2 years ago
8

How do you determine whether or not the paired t test is appropriate for a two-sample hypothesis?

Mathematics
1 answer:
Nat2105 [25]2 years ago
7 0

How to determine whether or not the paired t test is appropriate for a two-sample hypothesis is: The  two samples are independent.

<h3>What is Two-sample hypothesis?</h3>

Two-sample hypothesis can be defined as  a test that is conducted on two samples  so as to determine whether the two samples are independent.

A  paired t-test is appropriate when the scores are of  equal subject or  equal variances.

Therefore how to determine whether or not the paired t test is appropriate for a two-sample hypothesis is: The  two samples are independent.

Learn more about Two-sample hypothesis here:brainly.com/question/4232174

#SPJ1

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Four students each contribute $12 to buy a pizza, salad, and large soda. How much would each student pay if two more students al
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4 students contribute 12$ Each.

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Due to a manufacturing error, two cans of regular soda were accidentally filled with diet soda and placed into a 18-pack. Suppos
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Answer:

a) There is a 1.21% probability that both contain diet soda.

b) There is a 79.21% probability that both contain diet soda.

c)  P(X = 2) is unusual, P(X = 0) is not unusual

d) There is a 19.58% probability that exactly one is diet and exactly one is regular.

Step-by-step explanation:

There are only two possible outcomes. Either the can has diet soda, or it hasn't. So we use the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And \pi is the probability of X happening.

A number of sucesses x is considered unusually low if P(X \leq x) \leq 0.05 and unusually high if P(X \geq x) \geq 0.05

In this problem, we have that:

Two cans are randomly chosen, so n = 2

Two out of 18 cans are filled with diet coke, so \pi = \frac{2}{18} = 0.11

a) Determine the probability that both contain diet soda. P(both diet soda)

That is P(X = 2).

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 2) = C_{2,2}(0.11)^{2}(0.89)^{0} = 0.0121

There is a 1.21% probability that both contain diet soda.

b)Determine the probability that both contain regular soda. P(both regular)

That is P(X = 0).

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 0) = C_{2,0}(0.11)^{0}(0.89)^{2} = 0.7921

There is a 79.21% probability that both contain diet soda.

c) Would this be unusual?

We have that P(X = 2) is unusual, since P(X \geq 2) = P(X = 2) = 0.0121 \leq 0.05

For P(X = 0), it is not unusually high nor unusually low.

d) Determine the probability that exactly one is diet and exactly one is regular. P(one diet and one regular)

That is P(X = 1).

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 1) = C_{2,1}(0.11)^{1}(0.89)^{1} = 0.1958

There is a 19.58% probability that exactly one is diet and exactly one is regular.

8 0
3 years ago
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