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.... . . . . . . . . . . . . . . . . . . . . . . . . . 9.5 Limitations and Common Misinterpretations of Hypothesis Testing . . . . . . . . . . 1 1 6 10 15 17 Stat 3011 Chapter 9 CHAPTER 9: HYPOTHESIS TESTS Motivating Example A diet pill company advertises that at least 75% of its customers lose 10 pounds or more within 2 weeks. You suspect the company of falsely advertising the beneﬁts of taking their pills. Suppose you take a sample of 100 product users and ﬁnd that only 5% have lost at least 10 pounds. Is this enough to prove your claim? What about if 72% had lost at least 10 pounds? Goal: 9.1 Elements of a Hypothesis Test 1. Assumptions 2. Hypotheses Each hypothesis test has two hypotheses about the population: Null Hypothesis (H0 ): Alternative Hypothesis (Ha ): 1 Stat 3011 Chapter 9 Diet Pill Example: Let p = true proportion of diet pill customers that lose at least 10 pounds. State the null and alternative hypotheses for the diet pill example. 3. Test Statistic Deﬁnition: Test Statistic A test statistic is a measure of how compatible the data is with the null hypothesis. The larger the test statistic, the less compatible the data is with the null hypothesis. Most test statistics we will see have the following form: What does a large value of |T | reﬂect? NOTE: 2 Stat 3011 Chapter 9 4. p-value The p-value helps us to interpret the test statistic. Deﬁnition: p-value Assume H0 is true. Then the p-value is the probability...

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...STAT 302 – Statistical Methods Lecture 8 Dr. Avishek Chakraborty Visiting Assistant Professor Department of Statistics Texas A&M University Using sample data to draw a conclusion about a population • Statistical inference provides methods for drawing conclusions about a population from sample data. • Two key methods of statistical inference: o o Confidence intervals Hypothesis tests (a.k.a., tests of significance) Hypothesis Testing: Evaluating the effectiveness of new machinery at the Bloggs Chemical Plant • Before the installation of new machinery, long historical records revealed that the daily yield of fertilizer produced by the Bloggs Chemical Plant had a mean μ = 880 tons and a standard deviation σ = 21 tons. Some new machinery is being evaluated with the aim of increasing the daily mean yield without changing the population standard deviation σ. Hypothesis Testing: Evaluating the effectiveness of new machinery at the Bloggs Chemical Plant Null hypotheses • The claim tested by a statistical test is called the null hypothesis. The test is designed to assess the strength of the evidence against the null hypothesis. Usually the null hypothesis is a statement of “no effect” or “no difference”, that is, a statement of the status quo. Alternative hypotheses • The claim about the population that we are trying to find evidence for is the alternative hypothesis. The alternative hypothesis is one-sided if it states that a parameter is larger than or...

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...manufacturing costs. Distinguish between product and period costs. Explain the difference between a merchandising and a manufacturing income statement. Indicate how cost of goods manufactured is determined. Explain the difference between a merchandising and a manufacturing balance sheet. Identify trends in managerial accounting. Questions 1, 2, 3 Brief Exercises 1 Do It! 1 Exercises 1 A Problems B Problems *2. 4, 5, 6, 7, 8 11, 12 2, 3 1 *3. 4, 5, 7 2 2, 3, 4, 5, 6 3, 4, 5, 7, 13 8, 12, 13, 14, 15, 17 1A, 2A 1B, 2B *4. 13 6 2 1A, 2A 1B, 2B *5. 9, 14 3A, 4A, 5A 3B, 4B, 5B *6. 15, 16, 17, 18 8, 10, 11 3 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 14, 15, 16, 17 3A, 4A, 5A 3B, 4B, 5B *7. 10, 19, 20, 21 9 3A, 4A 3B, 4B *8. 22, 23, 24 25, 26 4 18 *Note: All asterisked Questions, Exercises, and Problems relate to material contained in the appendix to the chapter. Copyright © 2012 John Wiley & Sons, Inc. Weygandt, Managerial Accounting, 6/e, Solutions Manual (For Instructor Use Only) 1-1 ASSIGNMENT CHARACTERISTICS TABLE Problem Number 1A 2A 3A Description Classify manufacturing costs into different categories and compute the unit cost. Classify manufacturing costs into different categories and compute the unit cost. Indicate the missing amount of different cost items, and prepare a condensed cost of goods manufactured schedule, an income statement, and a partial......

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...x | y | x-xbar | y-ybar | (x-xbar)(y-ybar) | 14 | 2000 | -0.3 | -813.466667 | 244.0400001 | 19 | 9980 | 4.7 | 7166.533333 | 33682.70667 | 20 | 1000 | 5.7 | -1813.466667 | -10336.76 | 9 | 1000 | -5.3 | -1813.466667 | 9611.373335 | 15 | 1560 | 0.7 | -1253.466667 | -877.4266669 | 14 | 6559 | -0.3 | 3745.533333 | -1123.66 | 20 | 800 | 5.7 | -2013.466667 | -11476.76 | 5 | 908 | -9.3 | -1905.466667 | 17720.84 | 14 | 2800 | -0.3 | -13.466667 | 4.0400001 | 21 | 1800 | 6.7 | -1013.466667 | -6790.226669 | 4 | 1325 | -10.3 | -1488.466667 | 15331.20667 | 3 | 1000 | -11.3 | -1813.466667 | 20492.17334 | 16 | 1100 | 1.7 | -1713.466667 | -2912.893334 | 18 | 3261 | 3.7 | 447.533333 | 1655.873332 | 13 | 10282 | -1.3 | 7468.533333 | -9709.093333 | 10 | 8422 | -4.3 | 5608.533333 | -24116.69333 | 10 | 5500 | -4.3 | 2686.533333 | -11552.09333 | 9 | 1178 | -5.3 | -1635.466667 | 8667.973335 | 14 | 1254 | -0.3 | -1559.466667 | 467.8400001 | 6 | 8 | -8.3 | -2805.466667 | 23285.37334 | 12 | 3740 | -2.3 | 926.533333 | -2131.026666 | 35 | 6000 | 20.7 | 3186.533333 | 65961.23999 | 15 | 300 | 0.7 | -2513.466667 | -1759.426667 | 20 | 500 | 5.7 | -2313.466667 | -13186.76 | 8 | 3000 | -6.3 | 186.533333 | -1175.159998 | 16 | 1200 | 1.7 | -1613.466667 | -2742.893334 | 20 | 4241 | 5.7 | 1427.533333 | 8136.939998 | 14 | 1589 | -0.3 | -1224.466667 | 367.3400001 | 7 | 1497 | -7.3 | -1316.466667 | 9610.206669 | 28 | 600 | 13.7 | -2213.466667 |......

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...1. 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. Relative mean weight = (-53.68 + 64.64)/2 = 5.48. (Already provided) S.D = 22.93 Calculation: X bar 1.96 (SD) at 95% Confidence : 5.48 + 1.96(22.93) = 50.42 : 5.48 - 1.96(22.93) = -39.46 Expressed as: (-39.46, 50.42) 2. Which of the following values from Table 1 tells us about variability of the scores in a distribution? a. 60.22 b. 11.94 c. 22.57* d. 53.66?” 3. Assuming that the distribution for General Health Perceptions is normal, 95% of the females’ scores around the mean were between what values? Round your answer to two decimal places.” _ X = 39.71; σ = 25.46 X bar ± 1.96 (SD) = 39.71 – 1.96(25.46) = -10.19; 39.71 + 1.96(25.46) = 89.61 Expressed as: (-10.19, 89.61) 4. Assuming that the distribution of scores for Pain is normal, 95% of the men’s scores around the mean were between what two values? Round your answer to two decimal places. _ X = 52.53; σ = 30.90 Calculation: X bar ± 1.96 (SD) 52.53 – 1.96(30.90) = -8.03 52.53 + 1.96(30.90) = 113.09 Expressed as: (-8.03, 113.09) 5. Were the body image scores significantly different for women versus men? Provide a rationale for your answer.” They were, considering that the mean score for......

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...FUNDAMENTALS OF NURSING * Think about it… * ABCDEFGHIJKLMNOPQRSTUVWXYZ * Is represented as 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 * Then: * H-A-R-D-W-O-R-K * 8+1+18+4+23+15+18+11=98% * K-N-O-W-L-E-D-G-E * 11+14+15+23+12+5+4+7+5=96 % * A-T-T-I-T-U-D-E * 1+20+20+9+20+21+4+5=100 % * And look how far the love of God will take you * L-O-V-E-O-F-G-O-D * 12+15+22+5+15+6+7+15+4=101 % NURSING * Is the extent to which an individual or group is able to realize aspirations and satisfy needs and change or cope with environment. It is the complete physical, mental and social well-being and not merely the absence of disease or infirmity. Major Nursing Goals * Promotive – an action or measures designed to support behavior conducive to health * Preventive – any actions or measures designed to protect individuals, families, groups, communities from harm to their health * Curative – any actions or measures designed to correct or remove disease or any illness. * Rehabilitative – any actions or measures designed to restore health and promote recovery from any alteration of health Roles and Responsibilities of Nurses: * Caregiver – helping clients promote, restore, and maintain dignity, health and wellness * Communicator/Helper – central role of nurses in identifying needs of the client * Educator/Teacher – involves nursing activities, health promotion, the primary concern * Counselor – providing emotional, intellectual and......

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...refunds and the size of their incomes. Thus he took a sample of ten families and look at X = annual family income in thousands, and Y = the percentage of the refund spent within three months of their receipt. Family income in thousands of $ Percentage of refund spent is in % | |Family Income |% Refund spent | | | | | | |in 3 months | | | | | |X |Y |X^2 |Y^2 |X*Y | |1 |63 |20 |3969 |400 |1260 | |2 |68 |18 |4624 |324 |1224 | |3 |59 |25 |3481 |625 |1475 | |4 |75 |11 |5625 |121 |825 | |5 |86 |8 |7396 |64 |688 | |6 |69 |14 |4761 |196 |966 | |7 |78 |10 |6084 |100 |780 | |8 |99 |8 |9801 |64 |792 | |9 |49 |29 ......

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...use 0.75. Plotting the Binomial Probabilities 1. Create plots for the three binomial distributions above. Select Graph > Scatter Plot and Simple then for graph 1 set Y equal to ‘one fourth’ and X to ‘success’ by clicking on the variable name and using the “select” button below the list of variables. Do this two more times and for graph 2 set Y equal to ‘one half’ and X to ‘success’, and for graph 3 set Y equal to ‘three fourths’ and X to ‘success’. Paste those three scatter plots below. Calculating Descriptive Statistics Open the class survey results that were entered into the MINITAB worksheet. 2. Calculate descriptive statistics for the variable where students flipped a coin 10 times. Pull up Stat > Basic Statistics > Display Descriptive Statistics and set Variables: to the coin. The output will show up in your Session Window. Type the mean and the standard deviation here. Mean: 4.600 Standard deviation: 1.429 Short Answer Writing Assignment – Both the calculated binomial probabilities and the descriptive statistics from the class database will be used to answer the following questions. 3. List the probability value for each possibility in the binomial experiment that was calculated in MINITAB with the probability of a success being ½. (Complete sentence not......

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...Promotional | 2 | 78.00 | Proprietary Card | Female | Married | 30 | 8 | Regular | 1 | 22.50 | Visa | Female | Married | 40 | 9 | Promotional | 2 | 56.52 | Proprietary Card | Female | Married | 46 | 10 | Regular | 1 | 44.50 | Proprietary Card | Female | Married | 36 | 11 | Regular | 1 | 29.50 | Proprietary Card | Female | Married | 48 | 12 | Promotional | 1 | 31.60 | Proprietary Card | Female | Married | 40 | 13 | Promotional | 9 | 160.40 | Visa | Female | Married | 40 | 14 | Promotional | 2 | 64.50 | Visa | Female | Married | 46 | 15 | Regular | 1 | 49.50 | Visa | Male | Single | 24 | 16 | Promotional | 2 | 71.40 | Proprietary Card | Male | Single | 36 | 17 | Promotional | 3 | 94.00 | Proprietary Card | Female | Single | 22 | 18 | Regular | 3 | 54.50 | Discover | Female | Married | 40 | 19 | Promotional | 2 | 38.50 | MasterCard | Female | Married | 32 | 20 | Promotional | 6 | 44.80 | Proprietary Card | Female | Married | 56 | 21 | Promotional | 1 | 31.60 | Proprietary Card | Female | Single | 28 | 22 | Promotional | 4 | 70.82 | Proprietary Card | Female | Married | 38 | 23 | Promotional | 7 | 266.00 | American Express | Female | Married | 50 | 24 | Regular | 2 | 74.00 | Proprietary Card | Female | Married | 42 | 25 | Promotional | 2 | 39.50 | Visa | Male | Married | 48 | 26 | Promotional | 1 | 30.02 | Proprietary Card | Female | Married | 60 | 27 | Regular | 1 | 44.50 | Proprietary Card | Female | Married | 54 | 28 | Promotional |......

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... what is the minimum sample size needed if the total width of the confidence interval must be less than 5 percentage points (i.e., the confidence interval should extend at most 2.5 percentage points above and below the sample proportion)? Source Top of Form * 375. * 264. * 1,498. * The answer cannot be determined from the information given. 17. In a survey of twelve Harbor Business School graduates, the mean starting salary was $93,000, with a standard deviation of $17,000. The 95% confidence interval for the average starting salary among all Harbor graduates is: Source Top of Form * [$83,382; $102,618]. * [$82,727; $103,327]. * [$82,199; $103,801]. * [$59,000; $127,000]. 18. In a survey of 53 randomly selected patrons of a shopping mall, the mean amount of currency carried is $42, with a standard deviation of $78. What is the 95% confidence interval for the mean amount of currency carried by mall patrons? Source Top of Form * [$39.1; $44.9]. * [$24.4; $59.6]. * [$21.0; $63.0]. * [$14.4; $69.6]. 19. A filling machine in a brewery is designed to fill bottles with 355 ml of hard cider. In practice, however, volumes vary slightly from bottle to bottle. In a sample of 49 bottles, the mean volume of cider is found to be 354 ml, with a standard deviation of 3.5 ml. At a significance level of 0.01, which conclusion can the brewer draw? Source Top of Form * The true......

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...CASES DISPOSED, APPEALED, AND REVERSED IN HAMILTON COUNTY COURTS Common Pleas Court Judge Fred Cartolano Thomas Crush Patrick Dinkelacker Timothy Hogan Robert Kraft William Mathews William Morrissey Norbert Nadel Arthur Ney, Jr. Richard Niehaus Thomas Nurre John O� Connor Robert Ruehlman J. Howard Sundermann Ann Marie Tracey Ralph Winkler Total Total Cases Disposed 3,037 3,372 1,258 1,954 3,138 2,264 3,032 2,959 3,219 3,353 3,000 2,969 3,205 955 3,141 3,089 43,945 Appealed Cases 137 119 44 60 127 91 121 131 125 137 121 129 145 60 127 88 1,762 Reversed Cases 12 10 8 7 7 18 22 20 14 16 6 12 18 10 13 6 199 Common Pleas Court disposed 43, 945 total cases. The total number of cases appealed was 1,762, and the total number of cases reversed was 199. In all cases disposed in the Common Pleas Court, the probability of cases could be appealed and reversed is? Total reversed cases/ Total appealed cases= 199/1762= 0.11294 or 11.2% Domestic Relations court disposed a total of 30,499 cases. The total number of cases appealed was 106, and the total number of cases reversed was 17. In all cases disposed in the Domestic Relations Court, the probability of cases could be appealed and reversed is: Total reversed cases/ Total appealed cases=17/106= 0.16037 or 16% Municipal Court disposed of a total of 108,464. The total number of cases they appealed was 500, and the total number of cases reversed was 104. In all cases disposed in the Municipal Court, the probability of cases could be...

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...1600 / 400 = −1.83 σ/ n Lower tail p-value is the area to the left of the test statistic Using normal table with z = -1.83: p-value =.0336 Using Excel: p-value = NORMSDIST(-1.83) = .0336 c. d. p-value ≤ .05, reject H0. Conclude the mean refund of “last minute” filers is less than $1056. Reject H0 if z ≤ -1.645 -1.83 ≤ -1.645, reject H0 23. a. b. t= x − μ0 s/ n = 14 − 12 4.32 / 25 = 2.31 Degrees of freedom = n – 1 = 24 Upper tail p-value is the area to the right of the test statistic Using t table: p-value is between .01 and .025 Using Excel: p-value = TDIST(2.31,24,1) = .0149 c. c. p-value ≤ .05, reject H0. With df = 24, t.05 = 1.711 Reject H0 if t ≥ 1.711 2.31 > 1.711, reject H0. 24. a. b. t= x − μ0 s/ n = 17 − 18 4.5 / 48 = −1.54 Degrees of freedom = n – 1 = 47 Because t < 0, p-value is two times the lower tail area 9-2 Hypothesis Tests Using t table: area in lower tail is between .05 and .10; therefore, p-value is between .10 and .20. Using Excel: p-value = TDIST(1.54,47,2) = .1303 c. d. p-value > .05, do not reject H0. With df = 47, t.025 = 2.012 Reject H0 if t ≤ -2.012 or t ≥ 2.012 t = -1.54; do not reject H0 36. a. z= p − p0 p0 (1 − p0 ) n = .68 − .75 .75(1 − .75) 300 = −2.80 Lower tail p-value is the area to the left of the test statistic Using normal table with z = -2.80: p-value =.0026 Using Excel: p-value = NORMSDIST(-2.80) = .0026 b. z= .72 − .75 .75(1 − .75) 300 = −1.20 Lower tail p-value is the area......

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...STAT 346/446 - A computer is needed on which the R software environment can be installed (recent Mac, Windows, or Linux computers are sufficient).We will use the R for illustrating concepts. And students will need to use R to complete some of their projects. It can be downloaded at http://cran.r-project.org. Please come and see me when questions arise. Attendance is mandatory. Topics covered in STAT 346/446, EPBI 482 Chapter 5 – Properties of a Random Sample Order Statistics Distributions of some sample statistics Definitions of chi-square, t and F distributions Large sample methods Convergence in probability Convergence in law Continuity Theorem for mgfs Major Theorems WLLN CLT Continuity Theorem Corollaries Delta Method Chapter 7 – Point Estimation Method of Moments Maximum Likelihood Estimation Transformation Property of MLE Comparing statistical procedures Risk function Inadmissibility and admissibility Mean squared error Properties of Estimators Unbiasedness Consistency Mean-squared error consistency Sufficiency (CH 6) Definition Factorization Theorem Minimal SS Finding a SS in exponential families Search for the MVUE Rao-Blackwell Theorem Completeness Lehmann-Scheffe Location and scale invariance Location and scale parameters Cramer-Rao lower bound Chapter 9 - Interval Estimation Pivotal Method for finding a confidence interval Method for finding the “best” confidence interval Large sample confidence......

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...Whitners Autoplex Prices Differ from the National Average Team D RES 341 April 18, 2012 Dr. Leroy Paul Whitners Autoplex Prices Differ from the National Average Intro Research Process Consumers purchase new vehicles for many reasons: practicality, functionality, or as a status symbol. Trying to fill that need, Whitners Autoplex is an auto dealer that sells import and domestic automobiles (Lind, Marchal & Wathen, 2008). Buyers, ages 20 to 59, have purchased vehicles at Whitners. Team D, after reviewing the provided data, knows that over the past couple years the car dealer’s average selling price for an automobile was $22,000. Team D’s null hypothesis states that the selling prices in the data set remain unchanged from the established $22,000 average. However, Team D has faith that its alternate hypothesis that the average has changed will prevail. The following will provide the purpose of the research, problem definition, research hypothesis, and a look ahead to the following weeks concerning the study. Using historic and current data, Team D intends to prove the $22,000 average price of a car has changed, and most likely increased due to many factors. If Team D’s alternate hypothesis proves true, then Whitners Autoplex has been charging far more less than surrounding, or even nationwide, auto dealers. Knowing the average cost of a vehicle is important to auto dealers because profit is what allows a business to prosper and......

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...active cropland, ____ acres woodland, and approximately ____ acres of necessary and related land, including access roads, field edges and land, some suitable for future production. b) Notwithstanding the foregoing, Lessor recognizes that Lessee may incur startup costs at the beginning of the Lease term related to improving the agricultural soils on the Premises or making other improvements to the Premises. Toward that end, the first annual Lease fee, due on ___________, will be reduced by $_____. c) All Lease fees shall be due according to the schedule described in paragraph (a). A penalty of 1.5% per month will be assessed on late payments for each 30-day period that such payments are past due. This equates to an annual rate of approximately 18% on past due payments. d) The Lease fee may be renegotiated for any Lease extension. e) Farm utilities are not included and such costs are to be borne by the Lessee. V. Permitted Uses and Use Restrictions a) Lessee is hereby permitted all normal activities associated with agricultural purposes including but not limited to: planting, cultivating and harvesting of crops, including perennial crops; application of soil amendments; pest and weed management, erection and management of temporary structures such as greenhouses, hoop houses, temporary fencing, irrigation systems, livestock sheds etc.; use, routine maintenance and storage of tools and equipment; post-harvest washing, cooling, sorting, and packing; keeping of bees for farm......

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