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Damm [24]
2 years ago
13

Lyte wishes to study speed of Reaction Time to press a button in response to the onset of a lamp. The independent variable (V) i

s the color of the light produced by the lamp (red, orange, yellow, green, or blue) Since only 10 participants are available, she elects to administer the IV within-subjects with all 10 participants being exposed to all five levels of the color variable.
Mathematics
1 answer:
Vaselesa [24]2 years ago
6 0

Complete question:

Dr. Lyte wishes to study speed of Reaction Time to press a button in response to the onset of a lamp. The independent variable (V) is the color of the light produced by the lamp (red, orange, yellow, green, or blue) Since only 10 participants are available, she elects to administer the IV within-subjects with all 10 participants being exposed to all five levels of the color variable. The order of the color of the light presentation is to be counterbalanced. Using concepts from the textbook, why would Dr. Lyte need to use counterbalancing in this scenario?

Answer:

Here,

Independent variable (IV) is: the color of the light produced by the lamp (red, orange, yellow, green, or blue)

We are also told only 10 participants are available.

All 10 participants are being exposed to all five levels of the color variable in the same order.

Counterbalancing is said to be a technique used when establishing task order. It helps prevent introduction if cofounding variables.

Dr. Lyte will need to use counterbalancing technique in this scenario because some of the participants may be unable to understand difference in similar colours. Example some participants may not be able to differentiate between orange and red when the red colour comes after orange.

But using counterbalancing technique, Dr. Lyte can avoid such an error.

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2 years ago
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(CO 3) On average, the parts from a supplier have a mean of 97.5 inches and a standard deviation of 6.1 inches. Find the probabi
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Answer:

C. Probability is 0.90, which is inconsistent with the Empirical Rule.

Step-by-step explanation:

We have been given that on average, the parts from a supplier have a mean of 97.5 inches and a standard deviation of 6.1 inches.

First of all, we will find z-score corresponding to 87.5 and 107.5 respectively as:

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

z=\frac{87.5-97.5}{6.1}

z=\frac{-10}{6.1}

z=-1.6393

z\approx-1.64

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

z=\frac{107.5-97.5}{6.1}

z=\frac{10}{6.1}

z=1.6393

z\approx 1.64

Now, we need to find the probability P(-1.64.

Using property P(a, we will get:

P(-1.64

From normal distribution table, we will get:

P(-1.64

P(-1.64

P(-1.64

Since the probability is 0.90, which is inconsistent with the Empirical Rule, therefore, option C is the correct choice.

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2 years ago
In the diagram below, Bonnie claims that ΔMLV ≅ ΔRLT.
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∠ M ≅ ∠ R: true

<span>VL ≅ LT: true 

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) The GPA of accounting students in a university is known to be normally distributed. A random sample of 20 accounting students
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Answer:

The 95% confidence interval for the mean GPA of all accounting students at this university is between 2.5851 and 3.2549

Step-by-step explanation:

We are in posessions of the sample's standard deviation. So we use the student's t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 20 - 1 = 19

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.95}{2} = 0.975([tex]t_{975}). So we have T = 2.0930

The margin of error is:

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In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 2.92 - 0.3349 = 2.5851

The upper end of the interval is the sample mean added to M. So it is 2.92 + 0.3349 = 3.2549

The 95% confidence interval for the mean GPA of all accounting students at this university is between 2.5851 and 3.2549

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