1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
kondaur [170]
4 years ago
6

A human resource manager wants to see if there is a difference in the proportion of minority applicants who get the job and the

proportion of Caucasian applicants who get the job. After reviewing the records, she came up with the following confidence interval for the difference between the proportions: [ − 0.23 , − 0.15 ] . The sample mean difference was -0.19. The error bound is defined by: Right Endpoint - Sample Mean Difference Find the error bound.
Mathematics
1 answer:
mash [69]4 years ago
6 0

Answer:

Error Bound = 0.04

Step-by-step explanation:

Whenever we want to estimate parameter from a subset (or sample) of the population, we need to considerate that your estimation won't be a 100% precise, in other words, the process will have a random component that prevents us from always making the exact decision.

With that in mind, the objective of a confidence interval is to give us a better insight of where we expect to find the "true" value of the parameter with a certain degree of certainty.

The estivamative of the true difference between proportions was -0.19 and the confidence interval was [-0.23 ; -0.15].

The question also defines the error bound, as the right endpoint of the confidence interval minus the sample mean difference, so it's pretty straight foward:

Error Bound = -0.15 -(-0.19) = -0.15 + 0.19 = 0.04

The interpretation of this would be that we expect that the estimative for the difference of proportions would deviate from the "true" difference about \pm 0.04 or 4%.

You might be interested in
PLEASE HELP MEE!!! MAX POINTS AND BRAINLIEST!!
Maurinko [17]
Question does not specify full graph need more information
6 0
3 years ago
Beatrice built about 1/3 of a sandcastle. Linda built 4/7 of the same castle. What fraction of the sandcastle did they build tog
wlad13 [49]

Answer    

Find out the what fraction of the sandcastle did they build together .

To prove

Let us assume that the fraction of the sandcastle  they build together be x .

As given

Beatrice\ built\ about\ \frac{1}{3}\ of\ a\ sandcastle.

Linda\ built\ \frac{4}{7}\ of\ the\ same\ castle.

Than the equation becomes

x = \frac{1}{3} +\frac{4}{7}

L.C .M of (3,7) = 21

solving

x = \frac{7 + 3\times4}{21}

x = \frac{19}{21}

Therefore the fraction of the sandcastle did they build together be

x = \frac{19}{21}

Hence proved



4 0
3 years ago
Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m
Elenna [48]

Answer:

Part a: <em>The probability of no arrivals in a one-minute period is 0.000045.</em>

Part b: <em>The probability of three or fewer passengers arrive in a one-minute period is 0.0103.</em>

Part c: <em>The probability of no arrivals in a 15-second is 0.0821.</em>

Part d: <em>The probability of at least one arrival in a 15-second period​ is 0.9179.</em>

Step-by-step explanation:

Airline passengers are arriving at an airport independently. The mean arrival rate is 10 passengers per minute. Consider the random variable X to represent the number of passengers arriving per minute. The random variable X follows a Poisson distribution. That is,

X \sim {\rm{Poisson}}\left( {\lambda = 10} \right)

The probability mass function of X can be written as,

P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots

Substitute the value of λ=10 in the formula as,

P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{{\left( {10} \right)}^x}}}{{x!}}

​Part a:

The probability that there are no arrivals in one minute is calculated by substituting x = 0 in the formula as,

\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}}\\\\ = {e^{ - 10}}\\\\ = 0.000045\\\end{array}

<em>The probability of no arrivals in a one-minute period is 0.000045.</em>

Part b:

The probability mass function of X can be written as,

P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots

The probability of the arrival of three or fewer passengers in one minute is calculated by substituting \lambda = 10λ=10 and x = 0,1,2,3x=0,1,2,3 in the formula as,

\begin{array}{c}\\P\left( {X \le 3} \right) = \sum\limits_{x = 0}^3 {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \\\\ = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^1}}}{{1!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^2}}}{{2!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^3}}}{{3!}}\\\\ = 0.000045 + 0.00045 + 0.00227 + 0.00756\\\\ = 0.0103\\\end{array}

<em>The probability of three or fewer passengers arrive in a one-minute period is 0.0103.</em>

Part c:

Consider the random variable Y to denote the passengers arriving in 15 seconds. This means that the random variable Y can be defined as \frac{X}{4}

\begin{array}{c}\\E\left( Y \right) = E\left( {\frac{X}{4}} \right)\\\\ = \frac{1}{4} \times 10\\\\ = 2.5\\\end{array}

That is,

Y\sim {\rm{Poisson}}\left( {\lambda = 2.5} \right)

So, the probability mass function of Y is,

P\left( {Y = y} \right) = \frac{{{e^{ - \lambda }}{\lambda ^y}}}{{y!}};x = 0,1,2, \ldots

The probability that there are no arrivals in the 15-second period can be calculated by substituting the value of (λ=2.5) and y as 0 as:

\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = {e^{ - 2.5}}\\\\ = 0.0821\\\end{array}

<em>The probability of no arrivals in a 15-second is 0.0821.</em>

Part d:  

The probability that there is at least one arrival in a 15-second period is calculated as,

\begin{array}{c}\\P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\\\ = 1 - P\left( {X = 0} \right)\\\\ = 1 - \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = 1 - {e^{ - 2.5}}\\\end{array}

            \begin{array}{c}\\ = 1 - 0.082\\\\ = 0.9179\\\end{array}

<em>The probability of at least one arrival in a 15-second period​ is 0.9179.</em>

​

​

7 0
4 years ago
5. Find the change in temperature: from -1°F to -20°F
kap26 [50]

Answer:

-19

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Mariana is making a large pot of pasta. She mixes together 5 pounds of pasta, 6 pounds of sauce, and
Blizzard [7]

Answer:

B

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Other questions:
  • What other information do you need in order to prove the triangles
    12·1 answer
  • Juan recently entered into a contract for a cell phone plan. The table shows the monthly cost in dollars, f(x), if Juan sends x
    14·1 answer
  • Write 4.7 as a mixed number in simplest form
    9·1 answer
  • Lavonne sold 4 times as many raffle tickets as kenneth. lavonne sold 56 raffle tickets. how many tickets did kenneth sell?
    5·2 answers
  • D= 8 cm
    7·1 answer
  • If AC=10 inches and CB=5 inches what is AB
    10·2 answers
  • Find the approximate surface area of the cylinder below. Round your answer to the nearest tenth.
    11·2 answers
  • The area of the rectangle is 25xy^2, what is the width and length
    10·1 answer
  • Select the statement(s) that describe the plane(s) of symmetry for the solid. Select all that apply.
    8·1 answer
  • Identify the vertex of the function, F(x) =3(x-1)2 + 5.
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!