HLT362V Week 1 Homework EX#16 Answers for EXERCISE 16 page 122 (Questions 1- 4 are optional)• Mean and Standard Deviation Exercise 16: Mean and Standard Deviation 1. The researchers analyzed the data they collected as though it were at what level of measurement? a. Nominal b. Ordinal c. Interval/ratio d. Experimental Answer: c. The researchers analyzed the data as though it were at the interval/ratio level since they calculated means (the measure of central tendency that is appropriate
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□ EXERCISE 16 Questions to be Graded 1. The researchers analyzed the data they collected as though it were at what level of measurement? d. Experimental 2. What was the mean posttest empowerment score for the control group? The mean posttest empowerment score for the control group was 97.12 3. Compare the mean baseline and posttest depression scores of the experimental group. Was this an expected finding? Provide a rationale for your answer. The mean baseline depression score of
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Class: _Statistics____________Date: □ EXERCISE 16 Questions to be Graded 1. The researchers analyzed the data they collected as though it were at what level of measurement? d. Experimental 2. What was the mean posttest empowerment score for the control group? The mean posttest empowerment score was 97.12 3. Compare the mean baseline and posttest depression scores of the experimental group. Was this an expected finding? Provide a rationale for your answer. The mean baseline depression score
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Sohal‚ S 04/05/11 (HLT-362 V) Applied Statistics for Healthcare Professionals Exercise 18 Q1. Assuming that the distribution is normal for weight relative to the ideal and 99% of the male participants scored between ( - 53.68‚ 64.64)‚where did 95% of the values for weight relative to the ideal lie? Round your answer to two decimal places. Answer: Mean of weight relative to ideal = 5.48 and Standard Deviation (σ) = 22.93. Calculation: (x bar) 1.96(σ) 5.48± 1.96(22.93) 5.48 - 1.96(22.93)
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EXERCISE 36 6. Can ANOVA be used to test proposed relationships or predicted correlations between variables in a single group? Provide rationale for your answer. ANOVA cannot be used to test proposed relationships or predicted correlations between variables in a single group because it is designed to test for correlations and interactions amongst groups‚ i.e. in the test group of patients with OA you are testing the correlations between those who do not use GI and PMR and those that do. Although
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standard deviation were used to describe the length of labor. These were appropriate since mean and standard deviation can be calculated on an interval level of measurement. 3. Range could also be used to describe the length of labor since this statistic can be used on interval data with no natural zero point. 4. The distribution of scores was similar for the experimental and control groups for length of labor. The experimental group had a mean of 14.63 hours and the control group had a mean of
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PEARSON’S PRODUCT-MOMENT CORRELATION COEFFICIENT ANSWERS TO EXERCISE 23 Question 1 The r value for the relationship between Hamstring strength index 60o and the Shuttle run test is -0.149. This r value shows a weak correlation between the two variables‚ as it is less than the 0.3 threshold for significance. Therefore‚ the r value is not significant. Question 2 Between r=1.00 and r=-1.00‚ there is no difference in terms of strength. Both values are on the extreme ends of the spectrum and signify
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Applied Statistics at Grand Canyon University Exercise 20 from Workbook 1. Which patient scored the highest on the preoperative CVLT Acquisition? What was his or her T score? The 3rd patient scored 63 which is the highest CVLT T-score. 2. Which patient scored the lowest on postoperative CVLT Retrieval? What was this patient’s T score? The 4th patient scored the lowest on the postoperative CVLT Retrieval with a score of 23. 3. Did the patient in Question 2 have more of a memory
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Exercise 27 5. In the legend beneath Figure 2[->0]‚ the authors give an equation indicating that systolic blood pressure is SBP = 43.2 + 0.17x. If the value of x is postnatal age of 30 hours‚ what is the value for Ŷ or SBP for neonates ≤1‚000 grams? Show your calculations. Y = a + bx 43.2 + 0.17(30) = 48.3 The SBP for neonates ≤1‚000 grams at 30 hours is 48.3. 6. In the legend beneath Figure 2[->1]‚ the authors give an equation indicating that systolic blood pressure is SBP = 50.3 + 0.12x
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Exercise 36 Answers 1. Since the F value is significant‚ based on the p-value of 0.005 which is less than 0.05 which is sufficient to reject the null hypothesis. This suggests that there is a difference in the control and treatment groups. 2. Since the p- value is less than 0.05 and therefor the null hypothesis can be rejected. This presents that the mean‚ difficulty and mobility scores‚ must be different 3. The result was statistically significant with a probability score of p < 0.001.
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