This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis is an unformatted preview. Sign up to view the full document.
View Full DocumentMGMT 500 BUSINESS STATISTICS MIDTERM STUDY QUESTIONS Q1. The table shown below contains information technology (IT) investment as a percentage of total investment for eight countries during the 1990s. It also contains the average annual percentage change in employment during the 1990s. Explain how these data shed light on the question of whether IT investment creates or costs jobs. (Hint: Make use of relevant graphical tools) To analyze this we plot each countrys annual percentage change in employment during the 1990s on the vertical axis and the corresponding information technology (IT) investment as a percentage of total investment for eight countries on the horizontal axis. In order to sow the relationship, I created a scatter plot graph by the help of stat tools function of excel. As the points tend to move up and to the right, and as the correlation between these two variables is 0.930, this implies a reasonably strong positive linear relationship between IT investment and the jobs. Briefly we can say that IT investment creates jobs. Q2. 1 Country % IT % Change Netherlands 2.5% 1.6% Italy 4.1% 2.2% Germany 4.5% 2.0% France 5.5% 1.8% Canada 8.3% 2.7% Japan 8.3% 2.7% Britain 8.3% 3.3% U.S. 12.4% 3.7% for eight countries during the 1990s The percentage of the US population without health insurance coverage for samples from the 50 states and District of Columbia for both 2003 and 2004 produced the following tables of summary measures and correlations. Summary Measures Table: Percentage in 2003 Percentage in 2004 Count 51.000 51.000 Mean 14.455 14.855 Median 13.700 14.200 Standard deviation 3.724 4.098 Minimum 9.100 8.000 Maximum 24.900 26.300 First quartile 11.600 12.200 Third quartile 16.800 16.500 Skewness 0.910 0.699 Table of Correlations: Percent 2003 a. Describe the distribution of state percentages of Americans without health insurance coverage in 2004. Be sure to employ both measures of central location and dispersion in developing your characterization of this sample. ( 5 pts) The variance is essentially the average of the squared deviations from the mean. A more intuitive measure is the standard deviation, defined as the square root of the variance. The distribution of state percentages of Americans without health insurance coverage in 2004 can be summarized as below; As the mean of the population is 14.855; If the data was approximately symmetric, then the 1 st ,2 nd and the 3 rd quartiles had to include these below values; Approximately 68% of the observations are within 1 standard deviation of the mean, that is, within the interval XS which turns this range (10.757,18.953) Approximately 95% of the observations are within 2 standard deviation of the mean, that is, within the interval X2S which turns this range (6.659,23.051) Approximately 99.7% of the observations are within 3 standard deviation of the mean, that is, within the interval X3S which turns this range (2.561,27.149) As the distribution is skewed, we can say that in 2004, the state percentages of Americans ... View Full Document
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentPS 8
lecture 4-B
lecture10
Problem set 4B
Lecture 16
hw1dofile
Exam 1 Practice Questions Answer_ECON 305 (1)
termtestIIanswer_2013_S1
practice_questions_for_chapter_7_and_8 (1)
termtestIIanswer
HW 2
TEST1_Answer
Copyright © 2015. Course Hero, Inc.
Course Hero is not sponsored or endorsed by any college or university.