Am i right?

The true average diameter of ball bearings of a certain type is supposed to be 0.05 in. A one-sample t-test will be carried out

satela [25.4K]

Answer:

a) H0: $\mu \leq \mu_o$

H1: $\mu > \mu_o$

n = 13 represent the sample size

$t = 1.6$ represent the calculated statistic

The degrees of freedom are given by:

$df = n-1 = 13-1=12$

We can calculathe the p value with this formula:

$p_v = P(t_{12} >1.6) = 0.068$

Since $p_v >\alpha$ we fail to reject the null hypothesis on this case at 5% of significance.

b) H0: $\mu \geq \mu_o$

H1: $\mu < \mu_o$

n = 13 represent the sample size

$t = -1.6$ represent the calculated statistic

The degrees of freedom are given by:

$df = n-1 = 13-1=12$

We can calculathe the p value with this formula:

$p_v = P(t_{12}$

Since $p_v >\alpha$ we fail to reject the null hypothesis on this case at 5% of significance.

c) H0: $\mu \geq \mu_o$

H1: $\mu < \mu_o$

n = 25 represent the sample size

$t = -2.6$ represent the calculated statistic

The degrees of freedom are given by:

$df = n-1 = 25-1=24$

We can calculathe the p value with this formula:

$p_v = P(t_{24}$

Since $p_v$ we can reject the null hypothesis on this case at 1% of significance.

Step-by-step explanation:

Part a

For this case we assume that we are testing the following system of hypothesis

H0: $\mu \leq \mu_o$

H1: $\mu > \mu_o$

n = 13 represent the sample size

$t = 1.6$ represent the calculated statistic

The degrees of freedom are given by:

$df = n-1 = 13-1=12$

We can calculathe the p value with this formula:

$p_v = P(t_{12} >1.6) = 0.068$

Since $p_v >\alpha$ we fail to reject the null hypothesis on this case at 5% of significance.

Part b

For this case we assume that we are testing the following system of hypothesis

H0: $\mu \geq \mu_o$

H1: $\mu < \mu_o$

n = 13 represent the sample size

$t = -1.6$ represent the calculated statistic

The degrees of freedom are given by:

$df = n-1 = 13-1=12$

We can calculathe the p value with this formula:

$p_v = P(t_{12}$

Since $p_v >\alpha$ we fail to reject the null hypothesis on this case at 5% of significance.

Part c

For this case we assume that we are testing the following system of hypothesis

H0: $\mu \geq \mu_o$

H1: $\mu < \mu_o$

n = 25 represent the sample size

$t = -2.6$ represent the calculated statistic

The degrees of freedom are given by:

$df = n-1 = 25-1=24$

We can calculathe the p value with this formula:

$p_v = P(t_{24}$

Since $p_v$ we can reject the null hypothesis on this case at 1% of significance.

8 0

3 years ago

Write an equation for line M please and thank you!!!

Nikolay [14]

Answer:

y = -2x + 4

Step-by-step explanation:

First, find the slope by doing rise over run. I did this between the points (0, 4) and (2,0) This gives you -4/2 = -2.

Then, find the y-intercept, which is where the line goes through the y axis. This is at y = 4.

So, the equation is y = -2x + 4.

5 0

3 years ago

Read 2 more answers

What is the solution to the system x+3y+z=8 , 3x+2y-2z=5 , 4x-3t+kz=0

brilliants [131]

Answer:

Step-by-step explanation:

7 0

3 years ago

What is x in 15x + 25 = -20

serious [3.7K]

Hi. Good?

15x + 25 = -20
15x = -20 - 25
15x = -45
x = -45/15
x = -3

Thanks, I hope I have helped.

7 0

4 years ago

MAX POINTS

mrs_skeptik [129]

Let's do

$\\ \rm\dashrightarrow y=\dfrac{3}{x-h}+k$

Release k for some while

If

we take h=0

So

y=3/x

So vertical asymptote is at origin now

It mentioned that it's at x=-5 so we need to change x

put -5 in place of h

$\\ \rm\dashrightarrow y=\dfrac{3}{x-(-5)}$

$\\ \rm\dashrightarrow y=\dfrac{3}{x+5}$

Vertical asymptote at x=-5

Now

for k=0 horizontal asymptote at origin

But it's given

y is at 12

Same put y=12 in place of k

$\\ \rm\dashrightarrow y=\dfrac{3}{x+5}+12$

h=-5
k=12

Graph attached for verification

3 0

3 years ago