The standard deviation is expressed in absolute terms (i.e. in the same unit of measure as the data: lbs;inches;dollars;etc) while the coefficient of variation is relative measure? a. True. b. False.

a) True

Step-by-step explanation:

The given statement is true.

where are data points, is the mean and n is the number of observations.

The standard deviation express the variation of data from the mean and therefore, have the same unit as the data like lbs, inches, dollars, etc.

Coefficient of variation on the other hand is the ratio of standard deviation and mean. It is also a measure of dispersion but since it is a ratio, the units cancel each other.

Thus, coefficient of variation is dimensionless.

Thus, it is a relative measure.

Related Questions

What is the square rute of 81

The Square Root of 81 is 9.

Square roots are asking what number multiplied by itself equals the number inside the radical.

Find the repeating decimal between -2 1/3 and -2 1/5.

The repeating decimal in between -2 1/3 and -2 1/5 is - 2 4/15.

What is a number system?

The number system is a way to represent or express numbers.

A decimal number is a very common number that we use frequently.

Since the decimal number system employs ten digits from 0 to 9, it has a base of 10.

Any of the multiple sets of symbols and the guidelines for utilizing them to represent numbers are included in the Number System.

Given the mixed number,

-2 1/3 = - 7/3

Multiply and divide by 5

⇒ (-7/3) ×5/5 = -35/15

-2 1/5 = -11/5

Multiply and divide by 3

(-11/5) × 3/3 = -33/15

So in between -35/15 and -33/15 is -34/15 which is - 2.2666..

So,

-34/15 = -2 4/15

Hence "The repeating decimal in between -2 1/3 and -2 1/5 is - 2 4/15".

For more about the number system,

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let's first off, make the mixed fractions to improper fractions and then multiply one by the other's denominator, that way both fractions have the same denominator, so let's start,

Bank of America's Consumer Spending Survey collected data on annualcredit card charges in seven different categories of expenditures:transportation, groceries, dining out, household expenses, homefurnishings, apparel, and entertainment (U.S. AirwaysAttache, December 2003). Using data from a sample of 42 creditcard accounts, assume that each account was used to identify theannual credit card charges for groceries (population 1) and theannual credit card charges for dining out (population 2). Using thedifference data, the sample mean difference was = \$850, and the sample standard deviationwas sd = \$1,123. a.Formulate the null abd alternative hypothesis to test for no difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.
b.Use a .05 level of significance. Can you can conclude that the population mean differ? what is the p-value?
c. Which category, groceries or dining out, has a higher population mean annual credit card charge?What is the point estimate of the difference between the population means? What is the 95% confidence interval estimate of the difference between the population means?

H_o : u_d = 0 , H_1 : u_d ≠ 0

H_o rejected , p < 0.01

[ 499.969 < d < 1200.301 ] , d = 850

Step-by-step explanation:

Given:

- Difference in mean d = 850

- Standard deviation s = 1123

- The sample size n = 42

- Significance level a = 0.05

Solution:

- Set up and Hypothesis for the difference in means test as follows:

H_o : Difference in mean u_d= 0

H_1 : Difference in mean u_d ≠ 0

- The t test statistics for hypothesis of matched samples is calculated by the following formula:

t = d / s*sqrt(n)

Hence,

t = 850 / 1123*sqrt(42)

t = 4.9053

Thus, the test statistics t = 4.9053.

- The p-value is the probability of obtaining the value of the test statistics or a value greater.

Using Table 2, of appendix B determine p with DOF = n - 1 = 42 - 1 = 41 , We get:

p < 2*0.05 ----> 0.01

Thus,  p < 0.05  ....... Hence, H_o is rejected

- Set up and Hypothesis for the difference in means test as follows:

H_o : Difference in mean u_d =< 0

H_1 : Difference in mean u_d > 0

- The t test statistics for hypothesis of matched samples is calculated by te following formula:

t = d / s*sqrt(n)

Hence,

t = 850 / 1123*sqrt(42)

t = 4.9053

Thus, the test statistics t = 4.9053.

Using Table 2, of appendix B determine p with DOF = n - 1 = 42 - 1 = 41 , We get:

p < 0.005

Thus,  p < 0.05  ....... Hence, H_o is rejected

Hence, the point estimate is d = \$850

- The interval estimate of the difference between two population means is calculated by the following formula:

d +/- t_a/2*s / sqrt(n)

Where CI = 1 - a = 0.95 , a = 0.05 , a/2 = 0.025

Using Table 2, of appendix B determine p with DOF = n - 1 = 42 - 1 = 41 , We get:

t_a/2 = t_0.025 = 2.020

Therefore,

d - t_a/2*s / sqrt(n)

850 - 2.020*1123 / sqrt(20)

= 499.969

And,

d + t_a/2*s / sqrt(n)

850 + 2.020*1123 / sqrt(20)

= 1200.031

- The 95% CI of the difference between two population means is:

[ 499.969 < d < 1200.301 ]

a) Null hypothesis:

Alternative hypothesis:

b)

The next step is calculate the degrees of freedom given by:

Now we can calculate the p value, since we have a two tailed test the p value is given by:

4.905) =7.6x10^{-6}" alt="p_v =2*P(t_{(41)}>4.905) =7.6x10^{-6}" align="absmiddle" class="latex-formula">

So the p value is lower than any significance level given, so then we can conclude that we can to reject the null hypothesis that the difference between the two mean is equal to 0.

c) The confidence interval is given by:

For this case we have a confidence of 1-0.05 = 0.95 so we need 0.05 of the area of the t distribution with 41 df on the tails. So we need 0.025 of the area on each tail, and the critical value would be:

And if we find the interval we got:

We are confident 95% that the difference between the two means is between 499.96 and 1200.03

So we have enough evidence to conclude that one mean is higher than the other one, without conduct another hypothesis test because the confidence interval for the difference of means not contain the value of 0. And for this case the groceries would have a higher mean

Step-by-step explanation:

Previous concepts

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.

Part a

Let put some notation

x=test value for 1 , y = test value for 2

The system of hypothesis for this case are:

Null hypothesis:

Alternative hypothesis:

The first step is calculate the difference and we obtain this:

The second step is calculate the mean difference

This value is given

The third step would be calculate the standard deviation for the differences.

This value is given also

Part b

The 4 step is calculate the statistic given by :

The next step is calculate the degrees of freedom given by:

Now we can calculate the p value, since we have a two tailed test the p value is given by:

4.905) =7.6x10^{-6}" alt="p_v =2*P(t_{(41)}>4.905) =7.6x10^{-6}" align="absmiddle" class="latex-formula">

So the p value is lower than any significance level given, so then we can conclude that we can to reject the null hypothesis that the difference between the two mean is equal to 0.

Part c

The confidence interval is given by:

For this case we have a confidence of 1-0.05 = 0.95 so we need 0.05 of the area of the t distribution with 41 df on the tails. So we need 0.025 of the area on each tail, and the critical value would be:

And if we find the interval we got:

We are confident 95% that the difference between the two means is between 499.96 and 1200.03

So we have enough evidence to conclude that one mean is higher than the other one, without conduct another hypothesis test because the confidence interval for the difference of means not contain the value of 0. And for this case the groceries would have a higher mean

There are three children in a family. a friend is told that at least two of them are boys. what is the probability that all three are boys? the friend is then told that the two are the oldest two children. now what is the probability that all three are boys? use bayes law to explain this. assume throughout that each child is independently either a boy or a girl with equal probability.

It would be 33.333333% Likely to be a boy or a girl.

What is the difference? –2.03 – (–1.6)

–3.63

–0.43

0.43

3.63

I hope this helps you

-2.03+1.6

-0.43

If k(x) = 5x – 6, which expression is equivalent to (k + k)(4)? 05(4 + 4) - 6
O 5(5(4)-6) - 6
O 54 – 6 + 54 - 6
o 5(4) - 6 + 5(4) - 6​

k(x)=5x-6

k(4)=5(4)-6

(k+k)(4)=k(4)+k(4)

=5(4)-6+5(4)-6

I think.

What is the volume in cubic inches of a rectangular solid that is 6 feet long, 16 inches high, and 3 feet 6 inches wide?

(6*12)(16)(3*12+6)

72(16)42=48384 in^3
The volume is 130 cubic inches

Find the distance between (-3,4) and (1,7)

It will be sqrt((-3-1)^2 + (4-7)^2) = 5

Each paper clip can be traded for three matches. Each pencil can be traded for six paper clips.

Two pencils and two paper clips are worth (more than, less than, or equal to) thirty-eight matches.