12.6 pounds equals how many ounces

A study was recently conducted at a major university to estimate the difference in the proportion of business school graduates w

sveta [45]

Answer:

$(0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412$

$(0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318$

And the 95% confidence interval would be given (-0.1412;-0.0318).

We are confident at 95% that the difference between the two proportions is between $-0.1412 \leq p_A -p_B \leq -0.0318$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p_A$ represent the real population proportion for business

$\hat p_A =\frac{75}{400}=0.1875$ represent the estimated proportion for Business

$n_A=400$ is the sample size required for Business

$p_B$ represent the real population proportion for non Business

$\hat p_B =\frac{137}{500}=0.274$ represent the estimated proportion for non Business

$n_B=500$ is the sample size required for non Business

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$(0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412$

$(0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318$

And the 95% confidence interval would be given (-0.1412;-0.0318).

We are confident at 95% that the difference between the two proportions is between $-0.1412 \leq p_A -p_B \leq -0.0318$

7 0

3 years ago

2x + 3y = -1; 3x + 5y = -2

Andreas93 [3]

Answer:

x=1, y=-1. (1, -1).

Step-by-step explanation:

2x+3y=-1

3x+5y=-2

-----------------

3(2x+3y)=3(-1)

-2(3x+5y)=-2(-2)

-------------------------

6x+9y=-3

-6x-10y=4

--------------

-y=1

y=-1

2x+3(-1)=-1

2x-3=-1

2x=-1+3

2x=2

x=2/2

x=1

8 0

3 years ago

Which is the best estimate of 28x0.64

skad [1K]

Answer:

the full answer is 17.92, so that would round up to 18

Step-by-step explanation:

:)

3 0

3 years ago

I need help desperately

Nataly [62]

Answer:

B is the answer

Step-by-step explanation:

Look at where the decimal is placed and the value of the numbers

6 0

3 years ago

Which of the following expressions can be used to complete the equation below Secx / 1+cot^2x=

svlad2 [7]

Answer: D) Tan x sin x

<u>Step-by-step explanation:</u>

$\dfrac{sec}{1+cot^2}\\\\\\=\dfrac{1}{cos}\cdot \dfrac{1}{1+cot^2}\\\\\\=\dfrac{1}{cos}\cdot \dfrac{1}{1+(\frac{cos}{sin})^2}\\\\\\=\dfrac{1}{cos}\cdot \dfrac{1}{(\frac{sin^2}{sin^2})+(\frac{cos^2}{sin^2})}\\\\\\=\dfrac{1}{cos}\cdot \dfrac{sin^2}{sin^2+cos^2}\\\\\\=\dfrac{1}{cos}\cdot \dfrac{sin^2}{1}\\\\\\=\dfrac{sin\cdot sin}{cos\cdot 1}\\\\\\=\boxed{tan\cdot sin}$

3 0

3 years ago