tees that can be used for testing depending on the sample size and if the standard deviation is known. The t-test, which test of whether a sample observation comes from a larger sample with standard distribution of statistical properties. The z-test, “a statistical test used to determine whether two population means are different when the variances are known and the sample size is large.” (Investopedia, 2015) If the standard deviation is known then the z-test would be used, and when it is not known
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the amount of soda contained in each bottle. Upon receiving the previously mentioned information, I began my investigation by calculating the mean, median, and standard deviation for ounces inside of the randomly selected bottles. The mean for this data set is 14.87 ounces, the median for this data set is 14.8 ounces, and the standard deviation for this data set is 0.55 ounces (rounded to two decimal places). The next step in involved the problem is to constructing a 95% Confidence Interval for
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Chapter 13 13.5a Equal-variances estimator = (524 – 469) 2.009 = 55 73.87 b Equal-variances estimator = (524 – 469) 2.009 = 55 146.33 c The interval widens. d Equal-variances estimator = (524 – 469) 1.972 = 55 36.26 e The interval narrows. 13.6 = 0 0 a Equal-variances test statistic Rejection region: –2.074 or 2.074 = = .43, p-value = .6703. There is not enough evidence to infer that the population means differ. b Equal-variances test statistic
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emphasized the following five aspects when implementing the pay-for-performance system. 1. Differentiating employee’s performance The company strongly encouraged that all managers follow the normal distribution when assigning performance ratings to their direct reports. The motivation for the normal distribution was to encourage managers to honestly
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Median, and Standard Deviation of the sample size. * The Mean is the sum of the sample size in ounces divided by sample size. Based on the sample size of 30 bottles, there is a total of 446.1 ounces divided by the sample size of 30, gives is the mean of 14.87. = 14.87 * The Median is the data set middle number when the set is sorted in numerical order. Median of the 30 bottles is 14.8. * Standard deviation is the measure of how spread out the numbers is. To find the (Standard Deviation)
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Econometric Methods FIN5EME Semester 1, 2013 Assignment 2 * Cobb-Douglas cost function: TCi = µQiβ2 pi1β3 pi2β4 pi3β5 (1) Where, TCi= Total Cost for firm i Q= Output of firm i pi1= Wage Rate pi2= Rental Price of Capital pi3= Fuel Price * Taking the natural log of equation (1) log(TCi)= β1 + β2 log(Qi) + β3 log(pi1) + β4 log(pi2) + β5 log(pi3) + ei (2) where β1= (logµ) and ei= error term. * Eviews Output of the log-log model is as follows:
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means with an unknown standard deviation you use a t test and you use N-2 degrees of freedom. A. True | B. False | | Reset Selection Question 2 of 20 | 5.0 Points | Pretend you want to determine whether the mean weekly sales of soup are the same when the soup is the featured item and when it is a normal item on the menu. When it is the featured item the sample mean is 66 and the population standard deviation is 3 with a sample size of 23. When it is a normal item the sample mean is
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group that was an ideal age. * Is the random variable discrete or continuous? The variable is continuous there was a continuous number that could be named any age that the groups want to be as their ideal age. Question 2 Find the area under the normal curve that lies to the left of the following z-values. * Z=-1.30 * Z=-3.20 * Z=-2.56 *
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Oneway ANOVA Household Grocery Order Amounts $ Sum of Squares df Mean Square F Sig. Between Groups 29343.557 2 14671.778 7.128 .001 Within Groups 177019.882 86 2058.371 Total 206363.438 88 Post Hoc Test Multiple Comparisons Household Grocery Order Amounts $ Tukey HSD (I) Cities (J) Cities Mean Difference (I-J) Std. Error Sig. 95% Confidence Interval Lower Bound Upper Bound St. Catherines Edmonton -42.50460* 11.81485 .002 -70.6825 -14.3266 Moncton -9.23333 11.71430
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In the study, for application of one methodology author assumed that the returns are normally distributed and there is no cross-sectional dependence between stock returns. The simulation study for event days are implemented by assuming uniform distribution. Methodology and Data The methodology involves the use of CAPM model to estimate the returns of stocks. The stock portfolio is composed of 10, 25 and 50 stocks. Because of thin trade stock motivation the trade to trade adjustment was made. The
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