In testing the hypothesis H 0 colon mu subscript D less or equal than 5 space v s. space H 1 colon mu subscript D greater than 5 space, two random samples from two normal populations produced the following statistics: n subscript D equals 20 comma space X with bar on top subscript D equals 9 space a n d space s subscript D equals 7.5 . What conclusion can we draw at the 1% significance level?

The null hypothesis failed to be rejected (P-value=0.014).

Step-by-step explanation:

This is a hypothesis test for the population mean.

The null and alternative hypothesis are:

The significance level is α=0.01.

The sample has a size n=20.

The sample mean is M=9.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=7.5.

The estimated standard error of the mean is computed using the formula:

Then, we can calculate the t-statistic as:

The degrees of freedom for this sample size are:

This test is a right-tailed test, with 19 degrees of freedom and t=2.385, so the P-value for this test is calculated as (using a t-table):

As the P-value (0.014) is bigger than the significance level (0.01), the effect is  not significant.

The null hypothesis failed to be rejected.

answer: 1 + 1 = 11

step-by-step explanation:

option 3 . (5x + 1) - (-10x + 6)

step-by-step explanation:

option 1 :

(4x + 3) + (-9x + 4)

or 4x+3 -9x + 4

or -5x + 7

option 2

(x - 2) - (6x - 9)

or x-2-6x + 9

or -5x + 7

option 3

(5x+ + 6)

or 5x + 1 + 10x -6

or 15x -5

option 4

-4x+2x+7-3x

-7x + 2x + 7

-5x + 7

hence only option 3 did   not satisfy the given simplify expression.