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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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Romashka [77]

Answer:

A.The mean would increase.

Step-by-step explanation:

Outliers are numerical values in a data set that are very different from the other values. These values are either too large or too small compared to the others.

Presence of outliers effect the measures of central tendency.

The measures of central tendency are mean, median and mode.

The mean of a data set is a a single numerical value that describes the data set. The median is a numerical values that is the mid-value of the data set. The mode of a data set is the value with the highest frequency.

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  • Mean: If the outlier is a very large value then the mean of the data increases and if it is a small value then the mean decreases.
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  • Mode: The presence of outliers does not have any effect on the mode.

The mean of the test scores without the outlier is:

   \bar{x}=\frac{Total of the observations-Outlier value}{n-1} \\=\frac{(86*16)-72}{15} \\=\frac{1304}{15}\\ =86.9333

*Here <em>n</em> is the number of observations.

So, with the outlier the mean is 86 and without the outlier the mean is 86.9333.

The mean increased.

Since the median cannot be computed without the actual data, no conclusion can be drawn about the median.

Conclusion:

After removing the outlier value of 72 the mean of the test scores increased from 86 to 86.9333.

Thus, the the truer statement will be that when the outlier is removed the mean of the data set increases.

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