Bar Charts and Confidence Interval
The following data were collected on the examination results of biology students in 4 second level biology modules: There were 20 - 30 students in each class:
1. Vertebrate Physiology & Evolution:
56, 57, 67, 75, 45, 35, 56, 67, 82, 91, 58, 60, 61, 63, 76, 54, 36, 45, 67, 57, 60, 63, 76, 78, 50, 60, 54.
2. Practical Skills for Scientists
74, 67, 84, 67, 54, 89, 92, 95, 54, 67, 78, 78, 67, 73, 77, 80, 81, 85, 94, 87, 86
3. Cells & Sugars
56, 45, 34, 23, 45, 56, 67, 35, 45, 30, 28, 28, 35, 38, 42, 44, 58, 45, 34, 45, 57, 56, 32.
4. Introductory Microbiology
45, 67, 57, 55, 56, 59, 57, 59, 60, 64, 62, 63, 64, 64, 65, 63, 56, 58, 52, 53, 64, 65, 64, 53, 58, 55, 54, 62, 58, 60.
Plot these data with their associated confidence limits on a bar chart and write a paragraph what the data indicates.
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SOLUTION This solution is FREE courtesy of BrainMass!
The 95% confidence interval for the mean score in Vertebrate Physiology and Evolution is given as:
(x ̅-(z_(α/2) s)/√n,x ̅+(z_(α/2) s)/√n)
where,Sample Size,n=27
Mean,x Ì…=61.074
Standard Deviation,s=13.182
z_(α/2) for 95% confidence=1.96
Thus the 95% confidence interval for the mean score is given as
(61.074-(1.96×13.182)/√27,61.074+(1.96×13.182)/√27)
=(61.074-4.972,61.074+4.972)
=(56.102,66.046)
The 95% confidence interval for the mean score in Practical Skills for Scientists is given as:
(x ̅-(z_(α/2) s)/√n,x ̅+(z_(α/2) s)/√n)
where,Sample Size,n=21
Mean,x Ì…=77.571
Standard Deviation,s=11.788
z_(α/2) for 95% confidence=1.96
Thus the 95% confidence interval for the mean score is given as
(77.571-(1.96×11.788)/√21,77.571+(1.96×11.788)/√21)
=(77.571-0.765,77.571+0.765)
=(76.806,78.336)
The 95% confidence interval for the mean score in Cells & Sugars is given as:
(x ̅-(z_(α/2) s)/√n,x ̅+(z_(α/2) s)/√n)
where,Sample Size,n=23
Mean,x Ì…=42.522
Standard Deviation,s=13.182
z_(α/2) for 95% confidence=1.96
Thus the 95% confidence interval for the mean score is given as
(42.522-(1.96×11.630)/√23,42.522+(1.96×11.630)/√23)
=(42.522-4.753,42.522+4.753)
=(37.769,47.275)
The 95% confidence interval for the mean score in Introductory Microbiology is given as:
(x ̅-(z_(α/2) s)/√n,x ̅+(z_(α/2) s)/√n)
where,Sample Size,n=30
Mean,x Ì…=59.067
Standard Deviation,s=4.968
z_(α/2) for 95% confidence=1.96
Thus the 95% confidence interval for the mean score is given as
(59.067-(1.96×4.968)/√30,59.067+(1.96×4.968)/√30)
=(59.067-4.972,59.067+4.972)
=(54.095,64.039)
Please refer to the EXCEL sheet for the Bar Charts.
The given data indicates the scores on the respective modules. The average scores on each module are given by their mean values. The variance and standard deviation indicates how much the scores in each subject vary from their respective means. The greater the value of the variance or standard deviation, the higher is the amount by which the scores vary from their respective means. The scores in Vertebrate Physiology & Evolution show the highest variance. The scores in Introductory Microbiology show the least variance. The module Practical Skills for Scientists seems to be the easiest one, since the mean score is higher than the others. Also, the width of the confidence interval (indicated by the length of the line on top of the bar) is small as compared to the others.
https://brainmass.com/statistics/statistical-figures/bar-charts-confidence-interval-361352