Swati wants to distribute sweets to some of her friends.

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

2 years ago

A boat on the ocean is 2 mi from the nearest point on a straight? shoreline; that point is 15 mi from a restaurant on the shore.

Sladkaya [172]

Answer:

a) x = 0,70 miles

T = 1/2* (√4 + x²) +( 15 - x )/3 Objective Function to minimize

b) V(min) = 2.59 m/hr

Step-by-step explanation:

Let assume the boat is in Point A, she lands in point B, and R is the restaurant

Let call x distance between the point O in which perpendicular line from the boat get to shoreline, and the point were she land.

We know distance is

d = v*t ⇒ t = d/v

She rows at 2 miles/hr and walk at 3 miles /hr

According to that she takes

c (hypotenuse) of right triangle AOB/ 2 rowing

and 15 - x /3 walking

Total time is:

T = c/2 + ( 15-x )/3 c =√(2)² + x² ⇒ T = [√(2)² + x² ] /2 + ( 15-x )/3

T = 1/2* (√4 + x²) +( 15 - x )/3 (1)

And that is the Objective function to minimize

a) Taking derivatives on both sides of the equation we get

T´(t) = x /( √4 + x²) - 1/3 ⇒ T´(t) = 0 x /( √4 + x²) - 1/3 = 0

3*x - √( √4 + x²) = 0 ⇒ 3*x = √( √4 + x²)

Squaring both sides

9x² = 4 + x² ⇒ 8x² = 4 x² = 1/2 x = 0,70 miles

If we plug this value in the Objective function we will get the minimum time

1/2* (√4 + x²) +( 15 - x )/3 ⇒ [√ 4 + 0,5] /2 + 14,3/3 = T (min)

T(min) = 1.06 + 4.77 = 5.83 hr

Distance L (hypotenuse of right triangle AOR)

L = √(2)² + (15)² L = 15,13 miles

And that distance have to be traveled at least in 5.83 hr rowing

Then as v = d/t V(min) = 15.13/ 5.83 ⇒ V(min) = 2.59 m/hr

6 0

2 years ago

the ratio of the measures of the sides of a triangle is 21:8:14 if the perimeter of the triangle is 215 feet find the length of

SOVA2 [1]

Answer:

the 3 sides of the triangle are 105 ft, 40 ft and 70 ft.

Step-by-step explanation:

the perimeter is the sum of the 3 sides of the triangle

add the parts of the ratio 21 + 8 + 14 = 43

divide the perimeter by 43 to find the value of one part of the ratio

= 5 ft ← 1 part of the ratio, hence

21 parts = 21 × 5 = 105 ft

8 parts = 8 × 5 = 40 ft

14 parts = 14 × 5 = 70 ft

the 3 sides of the triangle are 105 ft, 40 ft and 70 ft

6 0

2 years ago

Frans filing cabinet is 6 feet tall 1 1/3 feet wide and 3 feet deep. She plans to paint all sides except the bottom of the cabin

aleksley [76]

Well you are look at a 3d image here
so if area is length times width then we need to find the area of every side of the cabinet and add it up
11/3 on the calculator is 3.66666667 sooooo....
top: 3 x 11/3 = 11
side 1: 3 x 6 = 18
side 2: 3 x 6 = 18
front: 6 x 11/3 = 22
back: 6 x 11/3 = 22
so we have all of our measurements:)
now we just add them up together.
11+18+18+22+22= 91!!
so Frans would have to paint 91 square inches of that cabinet:) hope i helped

6 0

2 years ago

the garza family drives to a state park. the remaining distance, in miles, to reach the state park is 285−60x , where x represen

alukav5142 [94]

285 - 60x ; where x represents the number of driving hours.

285 ⇒ The total distance to the state park.
60x ⇒ The number of miles driven after x hours.
60 ⇒ The number of miles driven after 1 hour.

y = 285 - 6x
y ⇒ The number of miles left to drive each day.

5 0

2 years ago

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