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statuscvo [17]
3 years ago
15

Pls help tyyyyyyyyyy

Mathematics
1 answer:
boyakko [2]3 years ago
7 0
You got this don't worry
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Andrea is told that the means of two groups in a study were statistically significant. She knows the means and standard deviatio
salantis [7]

Answer:

Cohen's D

Step-by-step explanation:

Cohen's D is a statistic that measures effect size. It shows standardised difference between 2 means.

Effect size is defined as how large the effect of a something is or its magnitude.

Cohen's D works effectively when the sample is >50 (that is for large samples). However a correction factor can be used to make results from small samples more accurate

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D = (mean1 - mean2) ÷ (√({standard deviation1}^2 + {standard deviation 2}^2)/2)

This is the most appropriate method in the given scenario

5 0
3 years ago
A stationary store has decided to accept a large shipment of ball-point pens if an inspection of 17 randomly selected pens yield
Leno4ka [110]

Answer:

0.762 = 76.2% probability that this shipment is accepted

Step-by-step explanation:

For each pen, there are only two possible outcomes. Either it is defective, or it is not. The probability of a pen being defective is independent from other pens. So we use the binomial probability distribution to solve this question.

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}.p^{x}.(1-p)^{n-x}

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

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

And p is the probability of X happening.

17 randomly selected pens

This means that n = 17

(a) Find the probability that this shipment is accepted if 10% of the total shipment is defective. (Use 3 decimal places.)

This is P(X \leq 2) when p = 0.1. So

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

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

P(X = 0) = C_{17,0}.(0.1)^{0}.(0.9)^{17} = 0.167

P(X = 1) = C_{17,1}.(0.1)^{1}.(0.9)^{16} = 0.315

P(X = 2) = C_{17,2}.(0.1)^{2}.(0.9)^{15} = 0.280

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.167 + 0.315 + 0.280 = 0.762

0.762 = 76.2% probability that this shipment is accepted

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What is the Lmc of 48 and 16
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