Ask question

Ask question

All categories

3 years ago

10

You want to compare the variation of heights of Econ 202 (population 1) and Econ 203 (population 2) students. You used a random

sample of 48 Econ 202 students and found a variance of 16 inches, while a sample of 63 Econ 203 students had a variance of 14 inches. If you test whether the variances of the two populations are equal, the test statistic will have a value of: Note. You should NOT change the way how the populations are labeled for this question. A. 0.9 B. 1.14 C. 1.56 D. 0.76

1 answer:

Ray Of Light [21]3 years ago

3 0

Answer:

Option B

Step-by-step explanation:

As per the question:

The variance of the 48 Econ 202, = 16 inches

The variance of 63 Econ 203 = 14 inches

Now, to test the variance of two populations:

For 48 Econ 202:

$s^{2} = 16$

For 63 Econ 203:

$s'^{2} = 14$

On comparing the two, the test statistics is given by;

F = $\frac{s^{2}}{s'^{2}}$

F = $\frac{16}{14} = 1.14$

You might be interested in

Suppose that the length of a side of a cube X is uniformly distributed in the interval 9

Nastasia [14]

Answer:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

Step-by-step explanation:

Given

$9 < x < 10$ --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

$v = x^3$

For a uniform distribution, we have:

$x \to U(a,b)$

and

$f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.$

$9 < x < 10$ implies that:

$(a,b) = (9,10)$

So, we have:

$f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Solve

$f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Recall that:

$v = x^3$

Make x the subject

$x = v^\frac{1}{3}$

So, the cumulative density is:

$F(x) = P(x < v^\frac{1}{3})$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$ becomes

$f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.$

The CDF is:

$F(x) = \int\limits^{v^\frac{1}{3}}_9 1\ dx$

Integrate

$F(x) = [v]\limits^{v^\frac{1}{3}}_9$

Expand

$F(x) = v^\frac{1}{3} - 9$

The density function of the volume F(v) is:

$F(v) = F'(x)$

Differentiate F(x) to give:

$F(x) = v^\frac{1}{3} - 9$

$F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}$

$F'(x) = \frac{1}{3}v^{-\frac{2}{3}}$

$F(v) = \frac{1}{3}v^{-\frac{2}{3}}$

So:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

8 0

3 years ago

1/2 + (-1/4)=????????????

erica [24]

Answer:

$\frac{1}{4}$

Step-by-step explanation:

$\frac{1}{2}$ = $\frac{2}{4}$ We need a common denominator before we can add.

$\frac{2}{4}$ + $\frac{-1}{4}$ means the same thing as

$\frac{2}{4}$ - $\frac{1}{4}$ = $\frac{1}{4}$

5 0

2 years ago

Please help me with this geometry question <br><br> image attached

vodomira [7]

15/17. The value (ratio) of cos A is 15/17.

The trigonometric ratios of an acute angle are, basically, the sine, the cosine and the tangent. They are defined from an acute angle, α, of a right triangle, whose elements are the hypotenuse, the leg contiguous to the angle, and the leg opposite the angle.

-The sine of the angle is the opposite leg divided by the hypotenuse.

-The cosine of the angle is the adjacent leg divided by the hypotenuse.

-The tangent of the angle is the opposite leg divided by the adjacent leg or, which is the same, the sine of the angle divided by the cosine of the angle.

cos A = adjacent leg/hypothenuse = BC/AC = 15/17

4 0

3 years ago

Convert 232°F to Celsius.<br><br> 55°C<br> 95°C<br> 100°C<br> 110°C

Ilya [14]

232- 449.6
55- 135
95- 203
100- 212
110- 230

3 0

3 years ago

Which is used to find the number of ways a given number of objects can be ordered?

bixtya [17]

Factorial is used to find the number of ways a given number of objects can be ordered

4 0

3 years ago

Read 2 more answers