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# Analyzing Cross-sectional Data and Correlation and Regression - Essay Example

## Extract of sample Analyzing Cross-sectional Data and Correlation and Regression

All this ratings indicated the level of customer satisfaction regarding our services.

The rating according to our instructors indicated that out of 100 hundred participants 12 rated the services as bad, 35 of them rated the services as neither bad or good, 39 rated the services as good and only 14 of them rated our services as very good, the majority therefore rated our services good. This can be graphically represented as follows

As graphically shown the mode is good, the graph is negatively skewed or skewed to the left meaning that the majority of the outcomes or observations are on the left of the graph, another notable observation is that none of the participants rated our instructors as very bad.

Our services were also rated through the quality of equipments used, out of 100 participants 6 of them rated the quality of equipments as very bad, 25 of them rated the equipments as bad, 33 of them rated the equipments as neither bad nor good, 27 of them rated the equipments as good and only 9 rated the equipments as very good.

Therefore the mode or the majority of the participants did not rate our services as bad or good regarding the quality of equipments, the mean of this observation was 3.08 and the standard error was 0.106059, the median and the mode were both 3, the results can be graphically represented as follows:

According to the participants our services were also rated according to the range of facilities av...
Therefore the mode or the majority of the participants did not rate our services as bad or good regarding the quality of equipments, the mean of this observation was 3.08 and the standard error was 0.106059, the median and the mode were both 3, the results can be graphically represented as follows:
quality of any equipment used
Frequency
6
25
33
Good
27
Very Good
9

0

We can construct a 95% confidence interval as follows
P(X - T Sx X + T Sx) = 95%
P ((3.08 - 0.172572 (0.106059)) (3.08 + 0.172572(0.106059)) = 95%
P ((3.0617) (3.183)) = 95%

According to the participants our services were also rated according to the range of facilities available, out of 100 participants only one rated the range of facilities available as very bad, 6 rated the facilities as bad, 20 as neither good or bad and 38 rated the facilities as good, the rest rated them as very good. Therefore according to the rating of the range of facilities we offer, 73 participants rated them as good or very good. This is graphically shown below:
The range of facilities available
Frequency
1
6
20
Good
38
Very Good
35

The mean was 4 and this shows that the average rated the range of equipments available as good; the mode was 4 which indicate that the majority of the participants rated the equipments as good. In the case where the mode, the median and mean are equal, the distribution assumes an asymmetric or bell shape where both deviations from the mean are identical, the negative value of skew ness indicates that the distribution is skewed to the left, the standard error of this observation was 0.094281 and ...Show more
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## Summary

This report describes the results of a research done on our members regarding the quality of services we offer on both full members and weekend members; it was aimed at exposing the level of our customer satisfaction regarding our services and equipment use…
Author : dallaszemlak
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