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3 years ago

9

DON'T DELETE MY QUESTION I WAS TRYING TO UPLOAD THE FILE YOU JUST INSTANTLY DELETED MY QUESTION b4 i could

1 answer:

lys-0071 [83]3 years ago

4 0

Answer:

The Distance should be 26

Step-by-step explanation:

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You want to estimate the mean amount of time college students spend on the Internet each month. How many college students must y

KonstantinChe [14]

Answer:

752.95

Step-by-step explanation:

Data provided in the question

The standard deviation of population = 210

The Margin of error = 15

The confidence level is 75%, so the z value is 1.96

Now the required sample size is

$= 1.96^2\times \frac{210^2}{15^2}$

= 752.95

Hence, the number of college students spends on the internet each month is 752.95

Simply we considered the above values so that the n could come

8 0

3 years ago

6squated - 3 squared

weeeeeb [17]

Answer:

27 is the answer to this equation

Step-by-step explanation:

6 times 6 equals 36

3times 3 equals 9

36

-9

27

7 0

3 years ago

the tax on a property with an assessed value of $64,000 is $650. Using a proportion, find the tax on a property with an assessed

AVprozaik [17]

The answer is $1200.

5 0

3 years ago

What is 19.2×10^8 minutes in hours

Nuetrik [128]

Hi!

$19.2*10^8$

$10^8=100,000,000$

$100,000,000*19.2$

$Answer-->=1,920,000,000$

Explanation: <u><em>This question is super easy. First you had to used 10^8 and it gave us 100,000,000. You can also multiply by 19.2*100,000,000 and it gave us the answer is 1,920,000,000 is the right answer. Hope this helps! And thank you for posting your question at here on Brainly. And have a great day. -Charlie</em></u>

6 0

3 years ago

Standard Error from a Formula and a Bootstrap Distribution Sample A has a count of 30 successes with and Sample B has a count of

tia_tia [17]

Answer:

Using a formula, the standard error is: 0.052

Using bootstrap, the standard error is: 0.050

Comparison:

The calculated standard error using the formula is greater than the standard error using bootstrap

Step-by-step explanation:

Given

Sample A Sample B

$x_A = 30$ $x_B = 50$

$n_A = 100$ $n_B =250$

Solving (a): Standard error using formula

First, calculate the proportion of A

$p_A = \frac{x_A}{n_A}$

$p_A = \frac{30}{100}$

$p_A = 0.30$

The proportion of B

$p_B = \frac{x_B}{n_B}$

$p_B = \frac{50}{250}$

$p_B = 0.20$

The standard error is:

$SE_{p_A-p_B} = \sqrt{\frac{p_A * (1 - p_A)}{n_A} + \frac{p_A * (1 - p_B)}{n_B}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.30 * (1 - 0.30)}{100} + \frac{0.20* (1 - 0.20)}{250}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.30 * 0.70}{100} + \frac{0.20* 0.80}{250}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.21}{100} + \frac{0.16}{250}}$

$SE_{p_A-p_B} = \sqrt{0.0021+ 0.00064}$

$SE_{p_A-p_B} = \sqrt{0.00274}$

$SE_{p_A-p_B} = 0.052$

Solving (a): Standard error using bootstrapping.

Following the below steps.

Open Statkey
Under Randomization Hypothesis Tests, select Test for Difference in Proportions
Click on Edit data, enter the appropriate data
Click on ok to generate samples
Click on Generate 1000 samples ---- <em>see attachment for the generated data</em>

From the randomization sample, we have:

Sample A Sample B

$x_A = 23$ $x_B = 57$

$n_A = 100$ $n_B =250$

$p_A = 0.230$ $p_A = 0.228$

So, we have:

$SE_{p_A-p_B} = \sqrt{\frac{p_A * (1 - p_A)}{n_A} + \frac{p_A * (1 - p_B)}{n_B}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.23 * (1 - 0.23)}{100} + \frac{0.228* (1 - 0.228)}{250}}$

$SE_{p_A-p_B} = \sqrt{\frac{0.1771}{100} + \frac{0.176016}{250}}$

$SE_{p_A-p_B} = \sqrt{0.001771 + 0.000704064}$

$SE_{p_A-p_B} = \sqrt{0.002475064}$

$SE_{p_A-p_B} = 0.050$

5 0

3 years ago