Question

2000: Mean= 6549/10 = 654.9 standard deviation= √(observation-mean) ^2/N-1 = √16697/9 = 43.07 2005: Mean =...

2000: Mean= 6549/10 = 654.9 standard deviation= √(observation-mean) ^2/N-1 = √16697/9 = 43.07 2005: Mean = 7030/10 = 703 standard deviation= √(observation-mean) ^2/N-1 = √ 8435/9 = 30.61 knowing this SD for each year did the average weekly earnings for the provinces become more or less variable over the period

Homework Answers

Answer #1

Result:

2000: Mean= 6549/10 = 654.9 standard deviation= √(observation-mean) ^2/N-1 = √16697/9 = 43.07

2005: Mean = 7030/10 = 703 standard deviation= √(observation-mean) ^2/N-1 = √ 8435/9 = 30.61

knowing this SD for each year did the average weekly earnings for the provinces become more or less variable over the period

coefficient of variation for 2000   = sd*100/mean = 43.07*100/654.9 = 6.576577

coefficient of variation for 2005 = 30.61*100/703 = 4.354196

By comparing the coefficient of variation for the two years, average weekly earnings for the provinces become less variable over the period.

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
.x is normally distributed with a mean of 27 and a standard deviation of 9. Take...
.x is normally distributed with a mean of 27 and a standard deviation of 9. Take a random sample of n = 9 observations and imagine that you have calculated a sample mean Xbar. What is the P(Xbar < 33) )? What is the population mean , sd value and Z score ? .x is normally distributed with a mean of 27 and a standard deviation of Take a random sample of n = 9 observations and imagine that you...
A variable is normally distributed with mean 10 and standard deviation 2. a. Find the percentage...
A variable is normally distributed with mean 10 and standard deviation 2. a. Find the percentage of all possible values of the variable that lie between 9 and 12. b. Find the percentage of all possible values of the variable that exceed 5. c. Find the percentage of all possible values of the variable that are less than 8.
A variable is normally distributed with mean 9 and standard deviation 2. a. Find the percentage...
A variable is normally distributed with mean 9 and standard deviation 2. a. Find the percentage of all possible values of the variable that lie between 4 and 12. b. Find the percentage of all possible values of the variable that exceed 7. c. Find the percentage of all possible values of the variable that are less than 8.
Study 1: the mean of 75 and a standard deviation of 10. Study 2: mean of...
Study 1: the mean of 75 and a standard deviation of 10. Study 2: mean of 75, standard deviation of 3. which has the higher probability for P(70<x<80)
1. For a population with a mean of μ = 55 and a standard deviation of...
1. For a population with a mean of μ = 55 and a standard deviation of σ = 24, what is the standard error of the distribution of sample means for a sample size of n=22? 2. If the population standard deviation is S= 17, how large a sample is necessary to have a standard error that is less than 10 points?
Find the mean and standard deviation for each binomial random variable: a. n = 40, ππ...
Find the mean and standard deviation for each binomial random variable: a. n = 40, ππ = .85 (Round your mean value to 2 decimal places and standard deviation to 4 decimal places.) Mean Standard deviation b. n = 78, ππ = .65 (Round your mean value to 2 decimal places and standard deviation to 4 decimal places.) Mean Standard deviation c. n = 30, ππ = .70 (Round your mean value to 2 decimal places and standard deviation to...
The standard deviation of the annual precipitations of a region over a period of 100 years...
The standard deviation of the annual precipitations of a region over a period of 100 years was 16 cm. During the last 15 years a standard deviation of annual precipitations was computed as 10 cm. Test the hypothesis that the precipitations in the region have become less variable than in the past, using a significance level of (a) 0.05 and (b) 0.01. Use chi-square distribution. . .
incubation period: mean = 8 days, standard deviation = 3 days illness period: mean = 4...
incubation period: mean = 8 days, standard deviation = 3 days illness period: mean = 4 days, standard deviation = 2 days 1. a. What is the probability that for a case load of 7 patients, their combined incubation period is less than 52 days? b. What is the probability that for a case load of 6 patients, their average incubation period is less than 6 days? c. What is the probability that for a case load of 7 patients,...
Calculate the mean and the (sample) standard deviation of the four numbers: 2, 3, 6, 9...
Calculate the mean and the (sample) standard deviation of the four numbers: 2, 3, 6, 9 1 (a) Two numbers, a and b, are to be added to this set of four numbers, such that the mean is increased by 1 and the (sample) standard deviation is increased by 2.5. Find the values of a and b. I want to the values of a and b
18) A distribution of random numbers has a mean or 10 and a standard deviation of...
18) A distribution of random numbers has a mean or 10 and a standard deviation of 1. The probability of a score below 9 is. A. 16% B.  84% 19) A distribution of random numbers has a mean or 10 and a standard deviation of 1. The probability of a score above 11 is A. 16% B. 33% 20) The vice president of marketing wants to issue a press release anytime monthly sales at on the top 5% for all months....
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT