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alexgriva [62]
3 years ago
10

6. A psychology professor of a large class became curious as to whether the students who turned in tests first scored differentl

y from the overall mean on the test. The overall mean score on the test was 75 with a standard deviation of 10; the scores were approximately normally distributed. The mean score for the first 20 students to turn in tests was 78. Using the .05 significance level, was the average test score earned by the first 20 students to turn in their tests significantly different from the overall mean?
Mathematics
1 answer:
grigory [225]3 years ago
6 0

Answer: Z is less than Zc ∴ 1.342 < 1.96

Therefore, Null hypothesis is not Rejected.

There is no sufficient evidence to claim that students turning in their test first score is significantly different from the mean.

Step-by-step explanation:

Given that;

U = 75

X = 78

standard deviation α = 10

sample size n = 20

population is normally distributed

PROBLEM is to test

H₀ : U = 75

H₁ : U ≠ 75

TEST STATISTIC

since we know the standard deviation

Z =  (X - U) / ( α /√n)

Z = ( 78 - 75 ) / ( 10 / √20)

Z = 1.3416 ≈ 1.342

Now suppose we need to test at ∝ = 0.05 level of significance,

Then Rejection region for the two tailed test is Zc = 1.96

∴ Reject H₀ if Z > Zc

and we know that Z is less than Zc ∴ 1.342 < 1.96

Therefore, Null hypothesis is not Rejected.

There is no sufficient evidence to claim that students turning in their test first score is significantly different from the mean.

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For a moving object, the force acting on the object varies directly with the object's acceleration. When a force of 20 N acts on
jeka94

The acceleration of the object will be 10 m/s²

Step-by-step explanation:

Direct variation is a relationship between two variables that can be expressed by an equation in which one variable is equal to a constant times the other

  • If y varies directly with x, then y ∝ x
  • y = k x, where k is the constant of variation

For a moving object, the force acting on the object varies directly with the object's acceleration.

Assume that the force is F and the acceleration is a

∵ F ∝ a

∴ F = k a

∵ F = 20 newtons

∵ a = 4 m/s²

- Substitute these values in the equation above to find k

∵ 20 = k (4)

∴ 20 = 4 k

- Divide both sides by 4

∴ k = 5

- Substitute the value of k in the equation

∴ F = 5 a ⇒ equation of variation

∵ F = 50 Newtons

∵ F = 5 a

∴ 50 = 5 a

- Divide both sides by 5

∴ 10 = a

∴ a = 10 m/s²

The acceleration of the object will be 10 m/s²

Learn more:

You can learn more about variation in brainly.com/question/10708697

#LearnwithBrainly

7 0
3 years ago
Translate and solve: What number is 283% of 1.89? Do not round.
zaharov [31]
I believe it should be 5.3487 since 283% is converted into 2.83 and multiplied by 1.89.
7 0
3 years ago
Identify the percent of change as an increase or a decrease. 50 pounds to 35 pounds
mojhsa [17]

Answer:

There is a 30% decrese.

Step-by-step explanation:

6 0
3 years ago
The mean percentange of a population of people eating out at least once a week is 57℅
Sidana [21]

Answer:

<u>The correct answer is B. between 56.45% and 57.55% </u>

Complete statement and question:

The mean percentage of a population of people eating out at least once a week is 57% with a standard deviation of 3.50%. Assume that a sample size of 40 people was surveyed from the population a significant number of times. In which interval will 68% of the sample means occur?

between 55.89% and 58.11%

between 56.45% and 57.55%

between 56.54% and 57.46%

between 56.07% and 57.93%

Source: brainly.com/question/1068489

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

Mean percentage of a population of people eating out at least once a week  = 57%

Standard deviation = 3.5%

Sample size = 40

Confidence level = 68%

2. In which interval will 68% of the sample means occur?

For answering this question, we should find out the standard deviation of the sample, using this formula:

Standard deviation of the sample = Standard deviation of the population/√Sample size

Standard deviation of the sample = 3.5/√40

Standard deviation of the sample = 3.5/6.32

Standard deviation of the sample = 0.55

Let's recall that a confidence level of 68% means that 68% of the sample data would have a value between the mean - 1 time the standard deviation of the sample and the mean  + 1 time the standard deviation of the sample. Thus:

57 - 1 * 0.55 = 57 - 0.55 = 56.45

57 + 1  * 0.55 = 57 + 0.55 = 57.55

<u>The correct answer is B. between 56.45% and 57.55% </u>

7 0
3 years ago
PLEASE HELP!!!
Scilla [17]
You need the Law of Cosines here and you use it when you have 2 sides and an enclosed angle.  The side across from the angle is the one we are looking for.  The correct way to express the Law using what we have is the last choice above.  Side RT is the unknown, and it is across from the angle that is enclosed between the 2 other sides.
3 0
3 years ago
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