Reprot And Analysis – Assignment Example

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The above table results can be represented by a histogram to achieve a clear visual representation of the results as follows: Fig 3.0: Total Survey ParticipationFrom the above representations it is clear that Australian population rated their health status as excellent or very good. Out of 950 participants, 52.4% (498 participants) believe their health is excellent or very good, 27.9% (265 participants) believes their health is good and 19.7% (187 participants) believe their health is fair or poor. Since the percentage of people with excellent or very good health is very high compared to other status, it is possible to conclude that most of the people in Australia are in very good health. Proportion of people in Australia who have Fair/Poor health, P = Therefore, P = 0.19795% confidence interval for this proportion can be calculated as follows: E = ZcZc = 1.96, P = 0.197, and n = 950Therefore, E = 1.96 E = 0.025Question 2This study is an experimental one because the subjects used in the study are subjected to different level of training but at the end of it they end up taking the same exam.

This means that the study was experimenting whether the training was contributing to the level of performance. If it were an observation study it would not have involved experimental manipulation, instead all subject could have been subjected to the same level of training and their scores collected and compare them to an earlier study. Mean = = = 45.96Median StemLeaf081720 4 6 7 7 8 930 2 2 3 4 5 5 6 8 8 41 2 4 6 6 8 9 9 50 0 1 2 2 3 8 8 960 0 1 3 5 771 1 4 88-93Median = {(n+1) ÷ 2th value= (47+1) ÷ 2= 24th valueFrom the cumulative frequency in the stem and leaf plot the 24th value is 46Therefore, the median value is 46Proportion obtaining at least a score of 40 = 28 ÷ 47= 0.596No training groupMean = = = 34Median8, 20, 26, 27, |30, 38|, 41, 49, 49, 52Median = (30 + 38) ÷ 2= 68 ÷ 2= 34Proportion obtaining at least a score of 40 = 4 ÷ 10= 0.4Stem and leaf plotStem Leaf081-20 6 730 8 41 9 9521 HR groupMean = 508 ÷ 10= 50.8Median 28, 32, 42, 44, |46, 48|, 52, 60, 63, 93= (46 + 48) ÷ 2= 47Proportion obtaining at least a score of 40 = 8 ÷ 10= 0.8Stem and leaf plotStem Leaf 283242 4 6 85260 3 932 HR groupMean= 543 ÷ 9= 60.33Median38, 50, 58, 59, |61|, 65, 67, 71, 74= 61Proportion obtaining at least a score of 40 = 8 ÷ 9= 0.89 Stem and leaf plotStem Leaf 384-50 8 961 5 7 71 45 HR groupMean = 388 ÷ 8= 48.5Median 32, 33, 35, |35, 46|, 58, 71, 78= (35 + 46) ÷ 2= 40.5Proportion obtaining at least a score of 40 = 4 ÷ 8= 0.5Stem and leaf plotStem Leaf 32 3 5 546 586-71 810 HR groupMean = 381 ÷ 10= 38.1 Median17, 24, 27, 29, |34, 36|, 50, 51, 53, 60 = (34 + 36) ÷ 2= 35Proportion obtaining at least a score of 40 = 4 ÷ 10= 0.4 Stem and leaf plot Stem Leaf 1724 7 9 34 6 50 1 3 60From the results obtained from b and c above the mean score of the whole group is 45.96 and the proportion of the group that scored at least 40 is 0.596.

This is an indication that more than half of the group scored at least 40 with or without training. The group which had no training performed poorly with a mean score of 34 and proportion that scored at least 40 was 0.4. This was an underperformance compared to the performance of the group.

The groups which were trained for 1 and 2 hours performed well with a mean score of 50.8 and 60.33 respectively. The proportion that scored at least 40 for the two groups was 0.8 and 0.89 respectively. The group that was training for 5 hours did not perform well and the group that was trained for 10 hours scored below the mean score. From the summary it is possible to conclude that the group required minimal training of about 2 hours to achieve best results.

Training the group for many hours seems to decrease their performance.

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