Let f(x)=2x-1, and g (x)=x^2-3

Researchers are investigating the effectiveness of leg-strength training on cycling performance. A sample of 7 men will be selec

ruslelena [56]

Answer:

A. The interval will be narrower if 15 men are used in the sample.

Step-by-step explanation:

Hello!

When all other things remain the same, which of the following statements about the width of the interval is correct?

A. The interval will be narrower if 15 men are used in the sample.

B. The interval will be wider if 15 men are used in the sample.

C. The interval will be narrower if 5 men are used in the sample.

D. The interval will be narrower if the level is increased to 99% confidence.

E. The interval will be wider if the level is decreased to 90% confidence.

Consider that the variable of interest "Xd: Difference between the peak power of a cyclist before training and after training" has a normal distribution. To construct the confidence interval for the population mean of the difference you have to use a pooled t-test.

The general structure for the CI is "point estimate"±" margin fo error"

Any modification to the sample size, sample variance and/or the confidence level affect the length of the interval (amplitude) and the margin of error (semiamplitude)

The margin of error of the interval is:

d= $t_{n-1;1-\alpha /2}$ * (Sd/n)

1) The sample size changes, all other terms of the interval stay the same.

As you can see the margin of error and the sample size (n) have an indirect relationship. This means, that when the sample size increases, the semiamplitude decreases, and when the sample size decreases, the semiamplitude increases.

↓d= $t_{n-1;1-\alpha /2}$ * (Sd/↑n)

↑d= $t_{n-1;1-\alpha /2}$ * (Sd/↓n)

Correct option: A. The interval will be narrower if 15 men are used in the sample.

2) The confidence level has a direct relationship with the semiamplitude of the interval, this means that when the confidence level increases, so do the semiamplitude, and if the level decreases, so do the semiamplitude:

↓d= ↓ $t_{n-1;1-\alpha /2}$ * (Sd/n)

↑d= ↑ $t_{n-1;1-\alpha /2}$ * (Sd/n)

I hope it helps!

4 0

3 years ago

Find the value of X round your number to the nearest tenths Explain it please

Anarel [89]

In this case, we must use trigonometry to find the value of x:

$tan(\alpha )=\frac{side-opposite-angle}{side-beside-angle} \\tan(70) = \frac{x}{3} \\3*tan(70) = x\\x = 3*tan(70)\\x = 8.24$

Thus the value of x rounded to the nearest tenth is 8.2

Hope that helps!

8 0

2 years ago

you had eight hours until you have to be on set if you will spend 3/4 of an hour driving to the TV station how much time does th

Inessa05 [86]

Answer:

7 hours and 15 minutes

Step-by-step explanation:

8 - 3/4 = 7 1/4 hours

7 0

3 years ago

Dos motos, A y B, toman la salida en una carrera de 90 km. La moto A realiza el recorrido con una velocidad media inferior en 20

aleksandr82 [10.1K]

Answer:

motorcycle A: 180 km/h

motorcycle B: 200 km/h

Step-by-step explanation:

To solve this question we need to write a system of equations for A and B, using the equation:

distance = speed * time

The speed of A is 20 less than the speed of B, so:

speedA = speedB - 20

And the time A traveled is 3 minutes (0.05 hours) more than B's time, so:

timeA = timeB + 0.05

Then, using the distance equation, we have that:

distanceB = speedB * timeB

90 = speedB * timeB

distanceA = speedA * timeA

90 = (speedB - 20) * (timeB + 0.05)

90 = speedB * timeB + 0.05 * speedB - 20*timeB - 1

90 = 90 + 0.05 * speedB - 20*timeB - 1

20*timeB = 0.05 * speedB - 1

timeB = (speedB - 20)/400

using this timeB in the distance equation, we have:

90 = speedB * (speedB - 20)/400

90 * 400 = speedB^2 - 20*speedB

speedB^2 - 20*speedB - 36000 = 0

Solving this quadratic equation, we have speedB = 200 km/h

And the speedA is:

speedA = speedB - 20 = 200 - 20 = 180 km/h

6 0

2 years ago