The number of hours per day that Americans spend on social networking is approximately normally distributed. A random sample of 20 Americans who use social networks had

B. (2.869, 3.411)

Step-by-step explanation:

The question is incomplete:

The number of hours per day that Americans spend on social networking is approximately normally distributed. A random sample of 20 Americans who use social networks had M = 3.14 hours and s = 0.58 hours. Find a 95% confidence interval for the actual mean number of hours that Americans spend on social networking.

A. (2.860, 3.420)

B. (2.869, 3.411)

C. (2.886, 3.394)

D. Not appropriate

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=3.14.

The sample size is N=20.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

The degrees of freedom for this sample size are:

The t-value for a 95% confidence interval and 19 degrees of freedom is t=2.093.

The margin of error (MOE) can be calculated as:

Then, the lower and upper bounds of the confidence interval are:

The 95% confidence interval for the mean is (2.869, 3.411).

answer is 144

step-by-step explanation:

answer:

it is general formula that is used to write value to its excet position

a11 a12 a13

a21 a22 a23

it is the example of order 2 by 3 matrix in this

a11 means value occur in 1 st row 1st column

a12 means value occurs in 1st row 2nd column

a13 means value occur in 1st row and 3rd column

a21 means value occurs in 2nd row and 1st column

a22 means value occurs in 2nd row and in 2nd column

a23 means value occurs in 2nd row and in 3rd column