Answer:
A sample of 801 is needed.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.
![\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
In which
z is the z-score that has a p-value of
.
The margin of error is of:
![M = z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
95% confidence level
So
, z is the value of Z that has a p-value of
, so
.
25% of U.S. homes have a direct satellite television receiver.
This means that ![\pi = 0.25](https://tex.z-dn.net/?f=%5Cpi%20%3D%200.25)
How large a sample is necessary to estimate the true population of homes which do with 95% confidence and within 3 percentage points?
This is n for which M = 0.03. So
![M = z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
![0.03 = 1.96\sqrt{\frac{0.25*0.75}{n}}](https://tex.z-dn.net/?f=0.03%20%3D%201.96%5Csqrt%7B%5Cfrac%7B0.25%2A0.75%7D%7Bn%7D%7D)
![0.03\sqrt{n} = 1.96\sqrt{0.25*0.75}](https://tex.z-dn.net/?f=0.03%5Csqrt%7Bn%7D%20%3D%201.96%5Csqrt%7B0.25%2A0.75%7D)
![\sqrt{n} = \frac{1.96\sqrt{0.25*0.75}}{0.03}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B1.96%5Csqrt%7B0.25%2A0.75%7D%7D%7B0.03%7D)
![(\sqrt{n})^2 = (\frac{1.96\sqrt{0.25*0.75}}{0.03})^2](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E2%20%3D%20%28%5Cfrac%7B1.96%5Csqrt%7B0.25%2A0.75%7D%7D%7B0.03%7D%29%5E2)
![n = 800.3](https://tex.z-dn.net/?f=n%20%3D%20800.3)
Rounding up:
A sample of 801 is needed.
Answer:
Check the answers to the questions below
Step-by-step explanation:
a) If
is the average learning rate of the 6th graders without background noise level and
is the average learning rate of the 6th graders with background noise level
The Null Hypothesis is that the learning rate of the 6th graders is not affected by the background noise level.
Null hypothesis, ![H_0: \mu_1 = \mu_2](https://tex.z-dn.net/?f=H_0%3A%20%5Cmu_1%20%3D%20%5Cmu_2)
b) The Alternative Hypothesis is that the learning rate of the 6th graders is affected by the background noise level.
Alternative hypothesis, ![H_a: \mu_1 \neq \mu_2](https://tex.z-dn.net/?f=H_a%3A%20%5Cmu_1%20%5Cneq%20%20%5Cmu_2)
<u>c) Assumptions that must be met about the data before she can correctly use independent t-test</u>
There must be random selection of the 6th graders
That the two groups are normally similar in their learning abilities
The division of students into the two groups should be at random
d) She has to make these assumptions to prevent bias and inaccuracy of results. If these assumptions are not made, the outcome of the experiment may not reflect the true effect of background noise on the learning of the 6th graders.
She can still get accurate results if she include some bias in the selection to prove a particular result.
Answer:
(B)
Step-by-step explanation:
Let us denote rectangular tiles by R and square tile by S, then
According to figure.
3 rectangular tiles +2 square tiles +2 square tiles =2 rectangular tiles +4 square tiles +1 rectangular tiles +4 square tiles
∴3R+2S+2R=2R+4S+R+4S
⇒5R+2S=3R+8S
⇒5R-3R=8S-2S
⇒2R=6S
⇒R=3S
Hence, 0ne rectangle =3 square
Answer:
A. 100$
Step-by-step explanation:
take 125 and multiply it to .20 (this is 20% in decimal form) this gives you 25. you know subtract 25 from 125 which gives you 100.