DATA EVALUATION
Question one
The scenario is taking place at NCLEX Memorial Hospital, within the Department of the Infectious Diseases Unit. In this department, there has been a rise in the number of patients admitted with a certain disease. The ages of the patients are playing an important role in the approached applied to treat them. In collaboration with the manager, statistical analysis has been used to examine closely at the patients’ ages. A spreadsheet of data having the number of the client, the status of infection disease, and age of the patient has been developed. The set of the data comprises of sixty patients that have the disease infection with their age within the age bracket of 35-76 yrs.
The qualitative variable of the study is the status of Infectious Disease. The quantitative variable is the number of patients and the ages of patients. The age of the patients is discrete variable, and the patient’s number is a continuous variable (Moore, 2008). The level of measurement of the patient’s number is ordinal can be used for a few statistical calculations. The level for measurement of the status of infectious disease is normal, and the level of measurement of the ages of the patients is interval since the variable can be used in various statistical calculations.
Question 2
The measure of the center can be described as the value that is at the middle or center of a set of data. A measure of the center includes mean, median and mode. A measure of the center is important because it provides an idea of the most normal, the most common or the representative feedback is going to be.
A measure of variation can be described as the way data is distributed. It is either feature of sample estimation of the date or the distribution probability of the data. It consists of range, variance, and standard deviation (Johnson & Bhattacharyya, 2010). It important since the measure of variation is important since it used to provide a comparison of variable among the set of data.
Question 3
- Mean is the average of the data provided. The mean of the data is (69+35+60+55+49+60+72+70+70+73+68+72+74+69+46+48+70+55+49+60+72+70+76+56+59+64+71+55+61+70+55+45+69+54+48+60+61+50+59+60+62+63+53+64+50+69+52+68+70+69+59+58+69+65+61+59+71+71+68) ÷ (60)
3654÷60= 60.9. This means that the average range of patients admitted with the infectious disease is 61 years.
- Median is the value in the middle. 35, 45, 46, 48, 48, 49, 49, 50, 50 ,52, 53, 54, 55, 55, 55, 55, 56, 58, 59, 59, 59, 59, 60, 60, 60, 60, 60, 61, 61, 61, 62, 63, 64, 64, 65, 68, 68, 69, 69, 69, 69, 69, 69, 69, 70, 70, 70, 70, 70, 70, 71, 71, 71, 71, 72, 72, 72, 73, 74, 76,
The media of the collected set of date is 61. In comparison to the mean, it shows that the ages of the patients is not skewed.
- The mode is the most often frequency of the set data. The above set of data the mode is 69. It the most often 69, 69, 69, 69, 69, 69, 69, repeating itself seven times. The most frequent patients admitted with the infectious disease is 69 years.
- The Midrange is the average between the minimum and maximum value. 35+76 ÷2= 56 years. It shows that the lowest age of patient admitted is 35 years and the highest is 76 years.
- Range is the highest value minus the least value. The range of the date set 76-35=41. It shows that the gap between the youngest patient with infectious disease and the oldest patient with that disease.
- The table shows the data used
| Number of the patient | Status of Infectious Disease | Ages of the patient (x) | Mean value (μ) | Square of mean deviation
(x-μ)² |
| 1 | yes | 69 | 61 | 64 |
| 2 | yes | 35 | 61 | 676 |
| 3 | yes | 60 | 61 | 1 |
| 4 | yes | 55 | 61 | 36 |
| 5 | yes | 49 | 61 | 144 |
| 6 | yes | 60 | 61 | 1 |
| 7 | yes | 72 | 61 | 121 |
| 8 | yes | 70 | 61 | 81 |
| 9 | yes | 70 | 61 | 81 |
| 10 | yes | 73 | 61 | 144 |
| 11 | yes | 68 | 61 | 49 |
| 12 | yes | 72 | 61 | 121 |
| 13 | yes | 74 | 61 | 169 |
| 14 | yes | 69 | 61 | 64 |
| 15 | yes | 46 | 61 | 225 |
| 16 | yes | 48 | 61 | 169 |
| 17 | yes | 70 | 61 | 81 |
| 18 | yes | 55 | 61 | 36 |
| 19 | yes | 49 | 61 | 144 |
| 20 | yes | 60 | 61 | 1 |
| 21 | yes | 72 | 61 | 121 |
| 22 | yes | 70 | 61 | 81 |
| 23 | yes | 76 | 61 | 225 |
| 24 | yes | 56 | 61 | 25 |
| 25 | yes | 59 | 61 | 4 |
| 26 | yes | 64 | 61 | 9 |
| 27 | yes | 71 | 61 | 100 |
| 28 | yes | 69 | 61 | 64 |
| 29 | yes | 55 | 61 | 36 |
| 30 | yes | 61 | 61 | 0 |
| 31 | yes | 70 | 61 | 81 |
| 32 | yes | 55 | 61 | 36 |
| 33 | yes | 45 | 61 | 256 |
| 34 | yes | 69 | 61 | 64 |
| 35 | yes | 54 | 61 | 49 |
| 36 | yes | 48 | 61 | 169 |
| 37 | yes | 60 | 61 | 1 |
| 38 | yes | 61 | 61 | 0 |
| 39 | yes | 50 | 61 | 121 |
| 40 | yes | 59 | 61 | 4 |
| 41 | yes | 60 | 61 | 1 |
| 42 | yes | 62 | 61 | 1 |
| 43 | yes | 63 | 61 | 4 |
| 44 | yes | 53 | 61 | 64 |
| 45 | yes | 64 | 61 | 9 |
| 46 | yes | 50 | 61 | 121 |
| 47 | yes | 69 | 61 | 64 |
| 48 | yes | 52 | 61 | 81 |
| 49 | yes | 68 | 61 | 49 |
| 50 | yes | 70 | 61 | 81 |
| 51 | yes | 69 | 61 | 64 |
| 52 | yes | 59 | 61 | 4 |
| 53 | yes | 58 | 61 | 9 |
| 54 | yes | 69 | 61 | 64 |
| 55 | yes | 65 | 61 | 16 |
| 56 | yes | 61 | 61 | 0 |
| 57 | yes | 59 | 61 | 4 |
| 58 | yes | 71 | 61 | 100 |
| 59 | yes | 71 | 61 | 100 |
| 60 | yes | 68 | 61 | 49 |
| ∑ | 4739 |
- Variance is the average deviation from the mean value. The total square mean variance is 4879, therefore, variance is ∑(X-μ)²/n =4739 ÷ 60=78.98
- The standard deviation is a tool of dispersion in the set of data. It is the square root of variance. For the set of data, it is√∑(X-μ) ²/n= √78.98=8.88. The value shows how age is dispersing.
This week you will begin working on Phase 2 of your course project. Using the same data set and variables for your selected topic, add the following information to your analysis:
- Discuss the importance of constructing confidence intervals for the population mean.
- What are confidence intervals?
- What is a point estimate?
- What is the best point estimate for the population mean? Explain.
- Why do we need confidence intervals?
- Based on your selected topic, evaluate the following:
- Find the best point estimate of the population mean.
- Construct a 95% confidence interval for the population mean. Assume that your data is normally distributed and ÃÆ’ is unknown.
- Please show your work for the construction of this confidence interval and be sure to use the Equation Editor to format your equations.
- Write a statement that correctly interprets the confidence interval in context of your selected topic.
- Based on your selected topic, evaluate the following:
- Find the best point estimate of the population mean.
- Construct a 99% confidence interval for the population mean. Assume that your data is normally distributed and ÃÆ’ is unknown.
- Please show your work for the construction of this confidence interval and be sure to use the Equation Editor to format your equations.
- Write a statement that correctly interprets the confidence interval in context of your selected topic.
- Compare and contrast your findings for the 95% and 99% confidence interval.
- Did you notice any changes in your interval estimate? Explain.
- What conclusion(s) can be drawn about your interval estimates when the confidence level is increased? Explain.
References
Johnson, R. A., & Bhattacharyya, G. K. (2010). Statistics: Principles and methods. Hoboken, NJ: John Wiley & Sons.
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Moore, D. S. (2008). The basic practice of statistics. New York: W.H. Freeman and Co.
PARTII
This week you will begin working on Phase 2 of your course project. Using the same data set and variables for your selected topic, add the following information to your analysis:
- Discuss the importance of constructing confidence intervals for the population mean.
- What are confidence intervals?
- What is a point estimate?
- What is the best point estimate for the population mean? Explain.
- Why do we need confidence intervals?
- Based on your selected topic, evaluate the following:
- Find the best point estimate of the population mean.
- Construct a 95% confidence interval for the population mean. Assume that your data is normally distributed and ÃÆ’ is unknown.
- Please show your work for the construction of this confidence interval and be sure to use the Equation Editor to format your equations.
- Write a statement that correctly interprets the confidence interval in context of your selected topic.
- Based on your selected topic, evaluate the following:
- Find the best point estimate of the population mean.
- Construct a 99% confidence interval for the population mean. Assume that your data is normally distributed and ÃÆ’ is unknown.
- Please show your work for the construction of this confidence interval and be sure to use the Equation Editor to format your equations.
- Write a statement that correctly interprets the confidence interval in context of your selected topic.
- Compare and contrast your findings for the 95% and 99% confidence interval.
- Did you notice any changes in your interval estimate? Explain.
- What conclusion(s) can be drawn about your interval estimates when the confidence level is increased? Explain.
| This assignment should be formatted using APA guidelines and a minimum of 2 pages in length. |
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