Using US Census data, a researcher compared home prices (in thousands of dollars) in the same102 Metropolitan Statistical Areas (MSAS) between 2004 and 2006. Her hypothesis is that the2006 prices will be less than the 2004 prices due to the collapse of the “bubble" in the housingmarket in 2005. The result of the analysis in Data Analysis is shown below:图2006 prices2004 pricesMean185.92210.40Variance12834.4917182.26Observationsdft StatP(T<=t) one-tailt Critical one-tailP(T<=t) two-tailt Critical two-tail1021025.15041E-172.36381.03008E-162.62549. Using the data output shown above, what conclusion can you draw from this analysis at a =.01?A.Home prices were significantly less in 2006 than in 2004; the Ho is rejected.B.There is no significant difference in time in home prices; the Ho is not rejected.C.Home pricas were significantly greater in 2006 than in 2004; the Họ is rejected.D.There is insufficient information for a decision on the Ho.

Answered: 2006 prices 2004 prices Мean 185.92…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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TB MC Qu. 11-16 (Algo) Consider the partial data set... Seved 52Fltiwrapper%252FitiaCloseW Help Consider the partial data set in the table represents online hours spent shopping by age and income. The average and standard deviation for the full data set is $47,713 and $14,269, respectively. Using z-scores to standardize the observations, what will be the average of standardized Income for the three provided? ID 2201 2202 Income 55,000 Online Age 56 Hours 2 50,000 50 2 2203 66,000 41 3 Multiple Choice O O 0.511 01690 11500 0.6510
QUESTION 4 We want to test whether males are taller than females. Heights of males and females are collected and the excel outputs are given below. Assume height is normally distributed for males and females with unequal variances. t-Test: Two-Sample Assuming Equal Variances t-Test: Two-Sample Assuming Unequal Variances Male Female Male Female Mean Variance Observations 170.86 169,1333 11.2249 15.93636 Mean 170.86 169.1333 Variance Observations Hypothesized Mean Difference 50 45 11.2249 15.93636 Pooled Variance 13.45398 50 45 Hypothesized Mean Difference df 93 86 t Stat 2.290934 t Stat 2.269888 P(T<=t) one-tail t Critical one-tail P(T<=t) one-tail. t Critical one-tail 0.012857 1.662765 0.012113 1.661404 P(T<=t) two-tail t Critical two-tail 0.025715 P(T<=t) two-tail t Critical two-tail 0.024226 1.985802 1.987934 You are given two excel outputs. Select the appropriate output to answer questions. Write down the p-value for this test given in the table (give you answer to 4 decimal…
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Determine the following: ● Mean Median Mode Standard deviation Variance Class interval 30-34 35-39 40-44 45-49 50 - 54 55-59 60-64 65-69 70-74 N f 3 6 7 12 10 5 4 2 1
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Table Physical Health Complaints and Burnout among Social Workers: One-Way Analysis of Variance Results Physical Health Complaints Low Burnout (n = 92) Middle Burnout (n = 93) High Burnout (n = 91) F(dfs) Respiratory infections M SD 5.7 2.3 7.0 3.0 8.8 3.7 23.8(2,273)* Overall Physical Health M SD 31.7 9.4 39.7 10.6 47.8 14.1 44.1(2,272)* * p < .05 Written Assignment Based on the information provided in the article excerpts and table, write a paragraph or two that describes and interprets the findings related to the one-way ANOVA presented in the table. Be sure to include sample size, relevant descriptive statistics, patterns in the data and an interpretation of the findings presented with supportive statistical information included. You do not need to talk about every number in the table, but you do need to include relevant statistical information to support your statements.
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Group 2 wieght gender 2. 37 2. 2 37 37 40 2. 2 2 40 40 43 gender Group 1 wieght 50 45 45 45 1 50 55 2. 2 46 47 1 58 60 60 47 48 2 2 48 48 1 1 62 49 50 50 1 68 2 68 69 69 1 2. 53 5B 2 58 60 60 70 Median Mode: Mean S Varinace Std. dev Median Mode: Mean Varinace Sd dev Sid Error. Std Error.
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2006 prices 2004 prices Мean 185.92 210.40 Variance 12834.49 17182.26 Observations 102 102 t Stat P(T