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stepladder [879]
2 years ago
9

2. When Professor Rodríguez conducted the proofreading study, the average number of errors detected in the print and computer co

nditions was 38.4 and 13.2, respectively; this difference was not statistically significant. When Professor Seuss conducted the same experiment, the means of the two groups were 21.1 and 14.7, but the difference was statistically significant. Explain how this could happen
Mathematics
1 answer:
Amanda [17]2 years ago
3 0

Answer:

Answer in explanation

Step-by-step explanation:

In this question, we are asked to determine what could have happened if the same experiment carried out by two academicians turned out to five different results statistically.

What majorly could have caused this is the difference in sample size. In a scientific experiment, the term sample size refers to how many individual observations are present.

Due to this difference in sample size, an experiment conducted with more sample size is expected to yield a greater amount of degree of freedom.

Thus, the small differences between the groups of observations are likely to be statistically significant.

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What is the true solution to the logarithmic equation below? log Subscript 4 Baseline left-bracket log Subscript 4 Baseline (2 x
Luda [366]

Step-by-step explanation:

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Step-by-step explanation:

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In 2009, a city with a population of 796,000 had 2,388 births. What is the number of births per 10,000 in the city?
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Connor borrows $8,000 at a rate of 19% interest per year. What is the amount due at the end of 7 years if the interest is compou
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Mopeds (small motorcycles with an engine capacity below 50 cm3) are very popular in Europe because of their mobility, ease of op
Mashcka [7]

Answer:

a) P(X<50)=0.9827

b) P(X>47)=0.4321

c) P(-1.5<z<1.5)=0.8664

Step-by-step explanation:

We will calculate the probability based on a random sample of one moped out of the population, normally distributed with mean 46.7 and standard deviation 1.75.

a) This means we have to calculate P(x<50).

We will calculate the z-score and then calculate the probability accordign to the standard normal distribution:

z=\dfrac{X-\mu}{\sigma}=\dfrac{50-46.7}{1.75}=\dfrac{3.7}{1.75}=2.114\\\\\\P(X

b) We have to calculatee P(x>47).

We will calculate the z-score and then calculate the probability accordign to the standard normal distribution:

z=\dfrac{X-\mu}{\sigma}=\dfrac{47-46.7}{1.75}=\dfrac{0.3}{1.75}=0.171\\\\\\P(X>47)=P(z>0.171)=0.4321

c) If the value differs 1.5 standard deviations from the mean value, we have a z-score of z=1.5

z=\dfrac{(\mu+1.5\sigma)-\mu}{\sigma}=\dfrac{1.5\sigma}{\sigma}=1.5

So the probability that maximum speed differs from the mean value by at most 1.5 standard deviations is P(-1.5<z<1.5):

P(-1.5

3 0
2 years ago
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