Please solve the following questions!

Question 3 A study was performed to determine whether men and women differ in their repeatability in assembling components on pr

Fed [463]

Answer:

There is no sufficient evidence to support the claim that men and women differ in repeatability for this assembly task

Step-by-step explanation:

Given

Let subscript 1 represent men and 2 represent women, respectively.

$n_1 = 25$

$n_2 = 21$

$s_1 = 0.98$

$s_2 = 1.02$

$\alpha = 0.02$

Required

Determine if here is enough evidence

First, we need to state the hypotheses

$H_o: \sigma_1^2 = \sigma_2^2$

$H_1: \sigma_1^2 \ne \sigma_2^2$

Next, calculate the test statistic using:

$F = \frac{s_1^2}{s_2^2}$

$F = \frac{0.98^2}{1.02^2}$

$F = 0.923$

Calculate the rejection region;

But first, calculate the degrees of freedom

$df_1 =n_1 - 1$

$df_1 =25 - 1$

$df_1 =24$

$df_2 = n_2 -1$

$df_2 = 21 - 1$

$df_2 = 20$

Using the F Distribution: table

$c = \frac{\alpha}{2}$

$c = \frac{0.02}{2}$

$c = 0.01$

At 0.01 level (check row 20 and column 24), the critical value is:

<em></em> $f_{0.01,24,20} = 2.86$ <em> --- the upper bound</em>

At 0.01 level (check row 24 and column 20), the critical value is:

$f_{0.01,20,24} = 2.74$

Calculate the inverse F distribution.

$f_{0.99,20,24} = \frac{1}{f_{0.01,20,24}} = \frac{1}{2.74} =0.365$ ---- the lower bound

The rejection region is then represented as:

$0.365 < Test\ Statistic < 2.86$

If the test statistic falls within this region, then the null hypothesis is rejected

$F = 0.923$ --- Test Statistic

$0.365 < 0.923 < 2.86$

<em>The above inequality is true; so, the null hypothesis is rejected.</em>

<em>This implies that, there is no sufficient evidence.</em>

7 0

3 years ago

How much is 7 kilometers

Ira Lisetskai [31]

7 kilometers=7000 meters

6 0

3 years ago

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Do you have a big brain if so need help plz!

tiny-mole [99]

Answer:

7x-15

4x-3

4x-5

7x +25

Step-by-step explanation:

6 0

3 years ago

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Geometry help. Inscribing circle help and constructing lines.

prohojiy [21]

<h3>1. Inscribe a circle ΔJKL. Identify the point of concurrency that is the center of the circle you drew. </h3>

To inscribe a circle in the attached triangle, we must follow the following steps. The resultant figure is the first one below.

1. Draw the angle bisector of angle JKL. The straight line in the green one.

2. Draw the angle bisector of angle KLJ. The straight line in the blue one.

3. Draw point D at these two lines.

4. Draw a line through point D perpendicular to side JK. The straight line in the red one.

5. Mark point G at the red line and side JK.

6. Draw a circle with center at D (point of concurrency) and a radius of DG.

<h3>2. Circumscribe a circle ΔFGH. Identify the point of concurrency that is the center of the circle you drew. </h3>

To circumscribe a circle in the attached triangle, we must follow the following steps:

1. Draw the perpendicular bisector of the side FG. The straight line in the red one. The resultant figure is the second one below.

2. Draw the perpendicular bisector of the side GH. The straight line in the blue one.

3. The center of the circle is the intersection of these two lines

4. Draw a circle with center at D (point of concurrency).

<h3>3. Construct the two lines tangents to circle</h3>

A tangent to a circle is a straight line that touches the circle at an only point. If a point A is outside a circle, we can draw two lines that pass through this point and are tangent to the circle at two different points each. So this is shown in the third figure. Lines red and blue are tangent to the circle at two different points.

3 0

3 years ago

Please help me on these attached files - The 1st one is the question and the others are answer choices - 10 POINTS WORTH!

Ganezh [65]

B, because it actually has the number of sponsors and is a positive slope.

8 0

3 years ago

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