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lara31 [8.8K]
2 years ago
11

An experiment was done to look at the positive arousing effects of imagery on different people. A sample of statistics lecturers

was compared against a group of students. Both groups received presentations of positive images (e.g., cats and bunnies), neutral images (e.g., duvets and light bulbs), and negative images (e.g., corpses and vivisection photographs). Positive arousal was measured physiologically (high values indicate positive arousal) both before and after each batch of images. The order in which participants saw the batches of positive, neutral and negative images was randomized to avoid order effects. It was hypothesized that positive images would increase positive arousal, negative images would reduce positive arousal and that neutral images would have no effect. Differences between the subject groups (lecturers and students) were not expected. What technique should be used to analyse these data? Three-way mixed ANOVA Two-way mixed ANOVA Three-way repeated-measures ANOVA Two-way mixed analysis of covariance

Mathematics
1 answer:
statuscvo [17]2 years ago
5 0

Answer:

The Three way mixed ANOVA

Step-by-step explanation:

The Three factor repeated measures ANOVA analysis otherwise referred to as the Three way mixed ANOVA should be used for the analysis.

This is because we are comparing three independent variables in the context.

• People type (statistic lectures and students)

• Imagery type (positive, neutral and negative imagery)

• Repeated measure: Before and After images shown

• Measuring the dependent variable positive arousal as shown in the table attached to the answer.

After going through the table attached you will see that there should be performed a three factor repeated measures ANOVA analysis.

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An ostrich can run 6 mph faster than a giraffe. An ostrich can run 7 miles in the same time that a giraffe can run 6 miles. Find
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Let t = time, s = ostrich, and g = giraffe.

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Answer:

The mean number of successful surgeries is 1.57.

The variance of the number of successful surgeries is 0.3111.

Step-by-step explanation:

STEP 1

If the tear on the left knee has a rim width of less than 3mm, the probability that the surgery on the left knee will be successful (ls) is 0.90.

That isP(Is)=0.90

.

The probability that the surgery on the left knee will fail (lf) is 0.10. That isP(V)=0.10

.

If the tear on the right knee has a rim width of 3-6 mm, the probability that the surgery on the right knee will be successful (rs) is 0.67. That isP(rs)=0.67

.

The probability that the surgery on the right knee will fail (rf) is 0.33.

That isP(vf)= 0.33

.

Let the random variable X denote the number of successful surgeries. The range of X is{0,1,2)

.

Now find the probabilities associated with the possible values of X. The number of successful surgeries is equal to 0 if the surgeries on both knees fail. Since the surgeries are independent, we have:

P(X = 0)=P(lf and rf)

= P()P(rf)

=(0.10)(0.33)

= 0.033

The number of successful surgeries is equal to 1 if the surgery on one knee is successful and the surgery on the other knee fails, that is

P(X =1)= P((ls and rf) or (if and rs))

= P(Is)P(rf)+P(V)P(rs)

= (0.90)(0.33)+(0.10)(0.67)

= 0.364

The number of successful surgeries is equal to 2 if the surgeries on both knees are successful, that is

P(X = 2)=P(ls and rs)

= P(Is) P(rs)

=(0.90)(0.67)

= 0.603

So, the probability mass function of X is the following

X        0              1               2

f(x)     0.033      0.364      0.603

The mean number of successful surgeries is,

E(X)=∑xP(x)

=(0×0.033) +(1×0.364)+(2×0.603)

=1.57

the mean number of successful surgeries is 1.57

The expected value is obtained by taking the summation of the product of each possible value of the random variable X with its corresponding probabilities. Thus, the mean number of successful surgeries is 1.57.

STEP 2

The variance of the number of successful surgeries is,

Var(X)=∑x²P(x)-(E(x²)

=(0×0.033) + (1 × 0.364) +(2 × 0.603) - (1.57)²

= 2.776 -(1.57)²

=0.3111

The variance of the number of successful surgeries is 0.3111.

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2 years ago
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