Pls explain deeply

A new anxiety medication has been manufactured and a study is being conducted to determine whether its effectiveness depends on

Irina18 [472]

The test statistics is used to determine if the anxiety medication can happen under a null hypothesis

The test statistics that is an appropriate hypothesis test is $z =\frac{0.35 - 0.40}{\sqrt{\frac{0.35(1 - 0.35)}{60} + \frac{0.40(1 - 0.40)}{85}}}$

<h3>How to determine the test statistic</h3>

<u>40 milligrams of medication</u>

Patients = 60

Lower stress level patients = 21

The mean of this medication is:

$\bar x = \frac{x}{n}$

So, we have:

$\bar x_1 = \frac{21}{60}$

$\bar x_1 = 0.35$

<u>75 milligrams of medication</u>

Patients = 85

Lower stress level patients = 34

The mean of this medication is:

$\bar x = \frac{x}{n}$

So, we have:

$\bar x_2 = \frac{34}{85}$

$\bar x_2 =0.4$

The test statistic is then calculated as:

$z =\frac{\bar x_1 - \bar x_2}{\sqrt{\frac{\bar x_1(1 - \bar x_1)}{n_1} + \frac{\bar x_2(1 - \bar x_2)}{n_2}}}$

The equation becomes

$z =\frac{0.35 - 0.40}{\sqrt{\frac{0.35(1 - 0.35)}{60} + \frac{0.40(1 - 0.40)}{85}}}$

Hence, the test statistics that is an appropriate hypothesis test is $z =\frac{0.35 - 0.40}{\sqrt{\frac{0.35(1 - 0.35)}{60} + \frac{0.40(1 - 0.40)}{85}}}$

Read more about test statistic at:

brainly.com/question/15980493

7 0

2 years ago

An object oscillates as it moves along the

Sauron [17]

Using it's formula, it is found that the displacement between t= 0 and t = 1 second is of 0 m.

The equation of motion is given by:

$x(t) = 4\sin{\left(\pi t + \frac{\pi}{6}\right)}$

The <em>displacement </em><u>between t= 0 and t = 1 second</u> is given by:

$d = x(1) - x(0)$

Hence:

$x(1) = 4\sin{\left(\pi + \frac{\pi}{6}\right)} = 4\sin{\left(\frac{7\pi}{6}\right)} = -\frac{4}{2} = -2$

$x(0) = 4\sin{\left(\frac{\pi}{6}\right)} = \frac{4}{2} = 2$

Then:

$d = x(1) - x(0) = -2 + 2 = 0$

The displacement is of 0 m.

You can learn more about displacement at brainly.com/question/248054

5 0

3 years ago

Help I need this done fast!

wolverine [178]

B

Sorry if I’m wrong...

5 0

3 years ago

A rectangle is 2m long. It’s diagonal is 3m long. Work out the width of the rectangle, to the nearest cm

Marizza181 [45]

Answer:

To the nearest cm

Width of the rectangle = 224cm

Step-by-step explanation:

Imagine a rectangle shape,

both opposite sides are equal, ie both lengths are equal and also applicable to the widths

Length is 2m

It's diagonal is 3m

The diagonal is a line that crosses from one length of a side to the other slantly, ie from one edge to the other, dividing the rectangle into two right angled triangles

The longest line will be the diagonal which is 3m

And the base length will be the length which is 2m

The width is what we are looking for which is represented with x

To solve this problem, we use Pythagoras' rule

x^2 = a^2 + b^2

Let x be diagonal side

a be length

b be the unknown width

Inserting into the formula

x^2 = a^2 + b^2

3^2 = 2^2 + b^2

9 = 4 + b^2

9 - 4 = b^2

5 = b^2

Square root of 5

b = 2.24m

The width = 2.24m

According to the question, we are asked to give the answer in nearest cm

Since 1cm = 0.01m

x cm = 2.24m

Then we cross multiply

1cm * 2.24m = xcm * 0.01m

x = 2.24 / 0.01

x = 224cm

5 0

3 years ago

Help with this problem.

ValentinkaMS [17]

15 is the answer I think because that is the Greatest Common Factor of all three of them.

I hope this helps!

5 0

3 years ago