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EleoNora [17]
3 years ago
8

Tessa conducts an experiment and obtains results that are statistically significant. What is meant by "statistically significant

"? O It means that Tessa used a large sample size. O It means that the results that Tessa obtained are too unusual to be explained by chance alone. It means that Tessa's experiment was not biased. O It means that the individuals in the experiment were randomly assigned to the treatment groups. earch 0 so DOLL​
Mathematics
1 answer:
Lesechka [4]3 years ago
6 0

9514 1404 393

Answer:

  (b)  It means that the results that Tessa obtained are too unusual to be explained by chance alone

Step-by-step explanation:

The test for statistical significance is a comparison of the result to the probability that it occurred by chance. The probability of a chance occurrence is usually set at 1% or 5%. The lower the percentage, the less likely a chance occurrence is, and the more difficult showing statistical significance becomes.

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Suppose that scores on a knowledge test are normally distributed with a mean of 60 and a standard deviation of 4.3. Scores on an
Sphinxa [80]

Question:

(c) Boris also took a logic test. His z-score on that test was +0.93 . Does this change the answer to which test Boris performed better on? Explain your answer using z-scores.

Answer:

The answers to the questions are;

(a) Based on the z score, Boris perform better on his aptitude test.

(b) Based on the z score, Callie perform better on his knowledge test .

(c) For Boris since +0.93 = z_{logic} >  z_{knowledge}  >  z_{aptitude}

Yes as Boris now performed best on the logic test.

Step-by-step explanation:

The z-score of a score is a measurement of the score withe respect to its distance from the mean as a factor of the standard deviation.

To solve the question, we note that we are required to find the z score as follows.

z score is given by z = \frac{x -\mu}{\sigma}

Where:

z = Standard score

x = Score

σ = Standard deviation

μ = Mean

(a) To find out which test did Boris performed better usin z score, we have

Boris scored a

57 on the knowledge test and

106 on the aptitude test

Therefore the z sore for the knowledge test is

z = \frac{x -\mu}{\sigma}

Here

x = 57

μ = 60

σ = 4.3

Therefore

z_{knowledge} = \frac{57 -60}{4.3} = -3/4.3 = -30/43 = -0.6977

The z sore for Boris on the aptitude test is

Here

x = 106

μ = 110

σ = 7.1

z_{aptitude} = \frac{106 -110}{7.1} = -40/17 = -0.5634

Based on the z score, Boris perform better on the aptitude test as his z score is higher (on the number line), --0.5634, compared to the z score on the knowledge test , -0.6977

(b) For Callie we have

Callie scored a

63 on the knowledge test and

114 on the aptitude test

Therefore the z sore for the knowledge test is

z = \frac{x -\mu}{\sigma}

Here

x = 63

μ = 60

σ = 4.3

Therefore

z_{knowledge} = \frac{63 -60}{4.3} = 3/4.3 = 30/43 = 0.6977

The z sore for Callie on the aptitude test is

Here

x = 114

μ = 110

σ = 7.1

z_{aptitude} = \frac{114 -110}{7.1} = 40/17 = 0.5634

Based on the z score, Callie perform better on the knowledge test as his z score is higher (on the number line), 0.6977, compared to the z score on the aptitude test , 0.5634.  

(c) If z_{logic}  = +0.93 then sinc for Boris z_{knowledge} = -0.6977 and

z_{aptitude}=  - 0.5634 then

z_{logic} >  z_{knowledge}  >  z_{aptitude}

Therefore Boris now performed best on the logic test.

6 0
3 years ago
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