NORTHCENTRAL UNIVERSITY ASSIGNMENT COVER SHEET Student: ODEDOYIN ABEL THIS FORM MUST BE COMPLETELY FILLED IN Follow these procedures: If requested by your instructor‚ please include an assignment cover sheet. This will become the first page of your assignment. In addition‚ your assignment header should include your last name‚ first initial‚ course code‚ dash‚ and assignment number. This should be left justified‚ with the page number right justified. For example: Student: ODEDOYIN A MGT
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Introduction: The purpose of the first experiment was to practice obtaining accurate and precise results by doing combustion analysis to find the amount of carbon dioxide and water released and the amount of fuel consumed. To identify the amount of carbon dioxide‚ water and burner consumed there were three equations used to find it. Equation 1 shows how much carbon dioxide was released. Mass of carbon dioxide = m2 - m1 Equation 1 was used as an experimental procedure created to help find out
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symbol in the book Steps: 1. Calculate Returns: Rt=Pt-Pt-1Pt-1= PtPt-1-1 2. Mean Monthly Returns and Standard Deviations for Each Stock: Mean Monthly Return: R= 1T*t=1TRt → AVERAGE Standard Deviation: STDEV = Var = 1T-1*t=1T(Rt-R)2 → STDEV.S 3. Convert Monthly to Annual Returns and Standard Deviations: Annual Mean Return: RA=12*RM Annual Standard Deviation: STDEVA= 12* VarM= 12*STDEVM 4. Portfolio (equally weighted): Monthly Return: RP= iwiRi equally weighted: RP=
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Web Calculator Exercise 1 Descriptive Statistics 1. The table below presents data for a sample of people who completed a religious survey. |Age |Gender |Denomination |Church Attendance | |56 |1 |7 |4 | |46 |2 |6 |5 | |49 |2 |6 |5 | |49 |1 |1
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distribution. In a positively skewed distribution the measures of central tendency will not be in the same place‚ they would be above the average‚ median‚ score. In an intelligence test where the mean is 100 and the standard deviation is 15 are distributed in a certain way. Standard deviation is a computed measure of how the scores vary around the mean score. This means that 100 is the average score of all the scores and the scores vary about 15 points around the mean score. If there are two groups‚ group
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process. Their client has offered samples to be analyzed‚ so they can quickly learn whether the process is operating satisfactorily or corrective actions needs to be taken. The numbers given in the case were as follows: assumed population standard deviation is equal to .21‚ sample size is equal to 30 and the test value of the mean was 12. They also stated the two hypotheses to be tested: the null hypothesis that the population is equal to 12 and the alternative hypothesis that the mean is not equal
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X has a normal distribution‚ and find the indicated probability. The mean is μ = 60.0 and the standard deviation is σ = 4.0. Find the probability that X is less than 53.0. Choose one answer. a. 0.5589 b. 0.0401 c. 0.9599 d. 0.0802 Question2 Marks: 1 Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation. Assume that the population has a normal distribution. Weights of eggs: 95% confidence; n = 22‚ x = 1
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Please read Article 1 and then answer the questions 1 - 4 45‚000 U.S. deaths to lack of insurance By Susan Heavey Susan Heavey – Fri Sep 18‚ 8:22 am ET. WASHINGTON (Reuters) – Nearly 45‚000 people die in the United States each year -- one every 12 minutes -- in large part because they lack health insurance and cannot get good care‚ Harvard Medical School researchers found in an analysis released on Thursday. "We’re losing more Americans every day because of inaction ... than drunk driving and
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Threesamma Joseph HLT-362V 9/7/2012 1. The answer is C‚ interval/ratio. The researchers analyzed the data as though it were interval/ratio level. They calculated the mean and standard deviation which is only appropriate for interval/ratio level data. 2. The mean post-test empowerment score for the control group is 97.12. This data is found explicitly in the chart of data given. 3. The baseline score mean is 14 and the post-test depression score mean is 13.36‚ meaning they were less depressed
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Effects of Caffeine and Nicotine on Lumbriculus variegatus INTRODUCTION An experiment was conducted to study and explore the circulatory system by exposing Lumbriculus variegatus‚ black worms‚ to household drugs. Lumbriculus variegatus was chosen as the experimental organism because of their transparent bodies and their simple physiology. Their transparent bodies help the experimenters to easily see their pulse. Another reason for choosing this specific organism is their body structure—large
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