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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30-year mortgage rate for small banks is less than 6 percent? Estimate the p-value. Chapter 11 27. A recent study focused on the number of times men and women who live alone buy take-out dinner in a month. The information is summarized below. Statistic Men Women Sample mean 24.51 22.69 Population standard deviation 4.48 3.86 Sample size 35 40 At the .01 significance level‚ is there a difference in the mean number of times men and women order take-out dinners in a month? What is the p-value
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1-2 Test 100 points In order to receive full credit‚ please show all work for every problem. You are welcome to use your calculator‚ book and notes‚ but please take this test on your own. 1. In your own words‚ explain what parameters and statistics are and what the difference is between them. Give an example to clarify. (10 points) 2. According to Consumer Reports‚ a random sample of 35 new cars gave an average of 21.1 mpg. What is the variable being studied? What is
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Chapter 11 Questions to be graded: 5. Were the experimental and control groups similar in their type of feeding? Provide a rationale for your answer. Answer: Yes the experimental and control groups are similar in their feeding type because 40.6% in the experimental group decided to breast-feed‚ and 41.7% in the control group decided to breast-feed‚ that’s only 1.1% difference. 53.1% of the experimental group bottle-fed while 50.0% of the control group bottle-fed‚ a small 3.1% difference. Finally
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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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QUESTION 21 The finishing process on new furniture leaves slight blemishes. The table below displays a manager’s probability assessment of the number of blemishes on one piece of new furniture. Number of Blemishes 0 1 2 3 4 5 Probability 0.34 0.25 0.19 0.11 0.07 0.04 1. On average‚ how many blemishes do we expect on one piece of new furniture? 2. What is the variance of blemishes on one piece of new furniture? (round to the nearest hundredth) QUESTION 22 The probability
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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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Practice Problems 1-KEY 1. The closing stock price of Ahmadi‚ Inc. for a sample of 10 trading days is shown below. Day Stock Price 1 84 2 87 3 84 4 88 5 85 6 90 7 91 8 83 9 82 10 86 For the above sample‚ compute the following measures. a. b. c. The mean = ∑X/n = 860/10 = 86 The median = (85+86)/2 = 85.5 The variance = ∑ X - X 2/ n-1 = {(84-86)2 + (87-86)2 + (84-86)2 + (88-86)2 + (85-86)2 + (90-86)2 + (91-86)2 + (83The standard deviation = √8.89 = 2.98 The coefficient of variation = 2.98/86 * 100%
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1. A density curve consists of a straight line segment that begins at the origin (0‚ 0) and has slope of 1. a. Sketch this density curve. What are the coordinates of the right endpoint of the segment? (Note that the right endpoint should be fixed so that the total area under the curve is 1. This is required for a valid density curve.) b. Determine the median‚ the first quartile (Q1)‚ and the third quartile (Q3). c. Relative to the median‚ where would you expect the mean of the distribution to lie
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