Given the following hypotheses: H0: μ = 490 H1: μ ≠ 490 A random sample of 15 observations is selected from a normal population. The sample mean was 495 and the sample standard deviation 9. Using the 0.01 significance level: a.) State the decision rule.
b.) Compute the value of the test statistic.
c.) What is your decision regarding the null hypothesis?

We conclude that the population mean is equal to 490.

Step-by-step explanation:

We are given that a random sample of 15 observations is selected from a normal population. The sample mean was 495 and the sample standard deviation 9.

Let = population mean.

So, Null Hypothesis, : = 490      {means that the population mean is equal to 490}

Alternate Hypothesis, : 490     {means that the population mean is different from 490}

The test statistics that will be used here is One-sample t-test statistics because we don't know about population standard deviation;

                               T.S.  =    ~

where, = sample mean = 495

            s = sample standard deviation = 9

             n = sample of observations = 15

So, the test statistics =    ~

                                     =  2.152

The value of t-test statistics is 2.152.

Now, at a 0.01 level of significance, the t table gives a critical value of -2.977 and 2.977 at 14 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as the test statistics will not fall in the rejection region.

Therefore, we conclude that the population mean is equal to 490.

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.