Download Azusa Pacific University: GNRS 613 midterm exam 1 (Answered). and more Exams Nursing in PDF only on Docsity!
Please answer each of the following questions.
1. For the data below, tell me the mean, median, mode, and standard deviation. The Data: 102.5, 104.5, 100.4, 95.9, 87, 95, 88.6, 89.2, 78.9, 84.6, 81.7, 72.2, 65.1, 68.1, 67.3, 52. The number of observations is 16 i. Mean = (102.5 + 104.5 + 100.4 + 95.9 + 87 + 95 + 88.6 + 89.2 + 78.9 + 84.
- 81.7 + 72.2 + 65.1+ 68.1 + 67.3 + 52.5)/16 = 83. ii. Median = (84.6+87)/2 = 85. iii. Mode = There is no mode amongst the given numbers iv. Standard deviation The formulae for calculating variance = Substituting the values: Variance = 212. GNRS 613: Graduate Statistics Midterm Exam Hint: Please review z scores ..... note that a score needs to be +/- 2 sd from the mean (or +/- a z score of 2) to be considered significant. There are a few questions on the exam related to this. I would like for you to work out the answers first.
Standard deviation = √ variance = √ 212. = 14.
2. For the data below, tell me the mean, median, mode, and standard deviation. 25, 24, 26, 31, 35, 38, 26, 24, 27, 26, 34, 36, 33, 31, 30 In this second question, we can use excel to solve it: i. Mean = 29. ii. Mode = 26 iii. Median = 30 iv. Standard deviation = 4. **For the three questions below…..think z scores and the area under the curve.
- A nursing professor was curious as to whether the students in a very large** class she was teaching who turned in their tests first scored differently from the overall mean on the test. The overall mean score on the test was 75 with a standard deviation of 10; the scores were approximately normally distributed. The mean score for the first 20 tests was 78. Did the students turning in their tests first score significantly different from the mean? Explain. Overall mean = 75 The standard deviation = 10 The mean score for the first 20 tests was 78 We are testing the mean of the first 20 tests, whether it is significantly different from the overall mean of the tests.
ability? Explain. In this case: The null hypothesis: Students from private universities do not have a high test-taking ability. Alternate hypothesis: The students from a private university have a higher test-taking ability. Because the standard deviation is known, we have to use z statistics. A significance level of 0.05 will be used in testing the hypothesis: In calculating for the test statistic, Z = Finding the p-value that corresponds to the z value: , we use the identity Eventually, the identity can be written as The calculated p-value is 0. This value exceeds the significance level of 0. Therefore, we can clearly state that the p-value is insignificant In the end, we can have the null hypothesis rejected. In conclusion, a student from a private university does not have a higher test-taking abil- ity.
5. Does computer-assisted instruction help community college health science students with reading difficulties to learn reading skills at a faster than normal rate? A researcher arranged for one of these students to have access
to a set of computer-learning programs instead of the normal reading curriculum for one term. At the end of the term, the researcher tested her on a standardized reading ability test on which the mean for students with reading difficulties is 36 with a standard deviation of 6. The test participant scored 55. What would you conclude? Explain. The mean is 36 The standard deviation is 6 The test participant scored 55 The test statistics: Substituting values t = (55-36)/6 = 19/6 = 3. The z value is 1.64, and the significance value is 0.05. This result shows an insufficient significance that computer-assisted instructions help stu- dents with reading difficulties to learn skills at a faster rate.
6. What is the normal curve and why is it so important to the world of inferential statistics? The normal curve refers to the spread of a variable following its normal distribution. The normal distribution is demarcated by a math formula. Most normal distributions often have different standard deviations (SD) and means. They are bell-curved, symmetrical and the greatest possible height is their mean. The mean, mode, and median are often
stood by all concerned parties. Overall, a good hypothesis may not conflict with natural laws often regarded as accurate and guarantee that available tools and techniques can be used in testing the relationship between variables. A null hypothesis depicts there is no statistical significance between two vari- ables. Thus, a null hypothesis illustrates that the results are obtained due to chance. The null hypothesis is the one the researcher attempts to disapprove. On the other hand, an al- ternative hypothesis depicts that the results are observed due to objective causes. The re- search hypothesis is that which the researcher attempts to prove. Consequently, a research hypothesis is a statement of prediction or expectation that will be evaluated by the re- searcher. Overall, research hypotheses are assumptions yet to be proved by scientific evi- dence.
Reference Hussain, J. (n.d.). Research hypothesis: Meaning, nature & importance. Maulana Mazharul Haque Arabic & Persian University. https://mmhapu.ac.in/doc/eCon- tent/Management/JamesHusain/Research%20Hypothesis%20-Meaning,%20Na- ture%20&%20Importance-Characteristics%20of%20Good%20%20Hypothesis %20Sem2.pdf
