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Fudgin [204]
3 years ago
13

We consider the differences between the reading and writing scores of a random sample of 200 students who took the High School a

nd Beyond Survey. The mean and standard deviation of the differences are −0.556 and 8.871 points.
Requried:
a. Calculate a 95% confidence interval for the average difference between the reading and writing scores of all students.
b. Interpret this interval in context. (Round your answers to two decimal places.)
Mathematics
1 answer:
Maslowich3 years ago
3 0

Answer:

A) True : Confidence Interval is the interval range around sample statistic, which is certain by extent of confidence level, to consist the actual population parameter.

B) True : Confidence Interval is the interval range around sample statistic, which is certain by extent of confidence level, to consist the actual population parameter.

C) False : Null Hypothesis can be accepted, despite of being actually false. This is called Type 2 Error.

Step-by-step explanation:

Hope this helps:)

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Solve the system by substitution.<br> y = -2<br> y =<br> 5x + 40
Colt1911 [192]

Answer:

x = 8.4

y = -2

Step-by-step explanation:

Step 1: Sub y=-2 into y=5x + 40

-2 = 5x + 40

Step 2: Solve for 'x'

-2 = 5x +40

-42 = 5x

x = 42/5

x = 8.4

Step 3: Solve for 'y'

y is given in the question, y=-2

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Step-by-step explanation:

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In a statewide lottery, one can buy a ticket for $1. With probability .0000001, one wins a million dollars ($1,000,000), and wit
Dahasolnce [82]

Using the expected value, it is found that the mean of the distribution equals $0.1.

  • The expected value, which is the mean of the distribution, is given by <u>each outcome multiplied by it's probability</u>.

The probabilities of <u>each outcome</u> are:

  • .0000001 probability of earning $1,000,000.
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Thus, the mean is given by:

E(Y) = 0.0000001(1000000) + 0.9999999(0) = 0.1

Thus showing that the expected value is $0.1.

A similar problem is given at brainly.com/question/24855677

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