What are the coordinates of the centroid of a triangle with vertices at D (0,0), E(-8,

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

1 year ago

the letters that spell Mississippi are each written on separate tiles lying face down on a table. a tile is selected at random,

Mars2501 [29]

Answer: The theoretical probability of choosing a tile with letter P =0.18

Step-by-step explanation:

Given word = MISSISSIPPI

Total number of letters in given the word = 11

Number of letter P in given word = 2

Let A be a event of choosing a tile with a letter P then

P(A) =Number of tiles with letter P / Total letters in given word

= 2 /11 = 0.18

6 0

1 year ago

If a cheeseburger weighs a half pound on Earth,<br> what will it weigh on Jupiter?

Katena32 [7]

Answer:

47 pounds i think \/'_'\/

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Peter buys tickets for his family to see his favorite movie, Elephant Mistake. He pays $37.50 for 2 adult tickets and 2 student

WARRIOR [948]

Answer:

the price of adult ticket and student ticket be $10.50 and $8.25 respectively

Step-by-step explanation:

Let us assume the price of adult ticket be x

And, the price of student ticket be y

Now according to the question

2x + 2y = $37.50

x + 3y = $35.25

x = $35.25 - 3y

Put the value of x in the first equation

2($35.25 - 3y) + 2y = $37.50

$70.50 - 6y + 2y = $37.50

-4y = $37.50 - $70.50

-4y = -$33

y = $8.25

Now x = $35.25 - 3($8.25)

= $10.5

Hence, the price of adult ticket and student ticket be $10.50 and $8.25 respectively

8 0

2 years ago

Just need to check these answers

amid [387]

Great Job! they are all correct. :)

Good luck in your next tests.

3 0

2 years ago