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3 years ago

14

What is the standard form of the equation 9(g-5h)

1 answer:

Lyrx [107]3 years ago

4 0

9g-45h, you have to distribute 9

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Sonbull [250]

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"You measure 34 dogs' weights, and find they have a mean weight of 67 ounces. Assume the population standard deviation is 13.5 o

liraira [26]

Answer:

The 95% confidence interval for the true population mean dog weight is between 62.46 ounces and 71.54 ounces.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{13.5}{\sqrt{34}} = 4.54$

The lower end of the interval is the sample mean subtracted by M. So it is 67 - 4.54 = 62.46 ounches.

The upper end of the interval is the sample mean added to M. So it is 67 + 4.54 = 71.54 ounces.

The 95% confidence interval for the true population mean dog weight is between 62.46 ounces and 71.54 ounces.

7 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

2 years ago

the remains of an ancient ball court include a rectangular playing alley with a perimeter of about 48 M. the length of the alley

Firdavs [7]

Answer:

Length is 20 m and width is 4 m.

Step-by-step explanation:

Given:

Perimeter of the rectangular alley, $P=48\textrm{ m}$

Length is 5 times the width.

Let width be $x$ .

So, as per question,

Length, $l = 5x$

Now, perimeter of rectangle is given as:

$P=2(l+b)$

Plug in 48 for $P$ , $5x$ for $l$ and $x$ for b.

$48=2(5x+x)\\48=2(6x)\\48=12x\\x=\frac{48}{12}=4$

Therefore, width is 4 m.

Length is $5x=5\times 4=20$ m.

8 0

3 years ago