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melamori03 [73]
3 years ago
11

Calculate the difference scores for the following data from a repeated measures study. Conduct a repeated measures t-test at apl

ha=.05 to find whether there was a change in scores. Subject A:Pre test score=34 post test score=39. Subject B:pre test score=41, post test score=48. Subject C:pre test score=38, post test score=35. Subject D: pre test score=29, post test score=36
Mathematics
1 answer:
Delicious77 [7]3 years ago
6 0

Answer:

There is no difference as per statistical evidence.

Step-by-step explanation:

We calculate t statistic from the formula

t =difference in means/Std error of difference

Here n1 = n2

t = (x bar - y bar)/sq rt of s1^2+s2^2

Let treatment I =X = 34 41 38 29

     Treatment II Y = 39 48 35 36

                     X       Y    

Mean          35.50   39.50

Variance     81.00   105.00

H0: x bar = y bar

Ha: x bar not equal to y bar

(Two tailed test at 0.05 significant level)

N1 = 4 and N2 = 4

df=N1+N2-2 = 6

s1^2 = 81/3 =27 and s2^2 = 105/3 = 35

Std error for difference =

t = -1.02

p =0.348834

p>0.05

Since p value >alpha we accept null hypothesis.

Hence there is statistical evidence to show that there is no difference in the mean level of scores.  

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Step-by-step explanation:

For each parachute, there are only two possible outcomes. Either there is damage, or there is not. The probability of there being damage on a parachute is independent of any other parachute, which means that the binomial probability distribution is used to solve this question.

To find the probability of damage on a parachute, the normal distribution is used.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

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C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

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Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Probability of a parachute having damage.

The opening altitude actually has a normal distribution with mean value 185 and standard deviation 32 m, which means that \mu = 185, \sigma = 32

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Z = \frac{X - \mu}{\sigma}

Z = \frac{100 - 185}{32}

Z = -2.66

Z = -2.66 has a p-value of 0.0039.

What is the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes?

0.0039 probability of a parachute having damage, which means that p = 0.0039

5 parachutes, which means that n = 5

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P(X \geq 1) = 1 - P(X = 0)

In which

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P(X \geq 1) = 1 - P(X = 0) = 1 - 0.9807 = 0.0193

0.0193 = 1.93% probability that there is equipment damage to the payload of at least one of five independently dropped parachutes.

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