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QveST [7]
3 years ago
14

You are testing the claim that the proportion of men who own cats is significantly different than the proportion of women who ow

n cats. You sample 80 men, and 55% own cats. You sample 100 women, and 30% own cats. Find the test statistic, rounded to two decimal places. 41.11Incorrect
Mathematics
1 answer:
otez555 [7]3 years ago
5 0

Answer:

Step-by-step explanation:

This is a test of 2 population proportions. Let 1 and 2 be the subscript for the men and women who own cats respectively. The population proportion of men and women who own cats would be p1 and p2 respectively.

p1 - p2 = difference in the proportion of men and women who own cats.

The null hypothesis is

H0 : p1 = p2

p1 - p2 = 0

The alternative hypothesis is

Ha : p1 ≠ p2

p1 - p2 ≠ 0

it is a two tailed test

Sample proportion = x/n

Where

x represents number of success

n represents number of samples

For men

n1 = 80

p1 = 55/100 = 0.55

x1 = p1n1 = 0.55 × 80 = 44

For women,

n2 = 100

p2 = 30/100 = 0.3

x2 = p2n2 = 0.3 × 100 = 30

The pooled proportion, pc is

pc = (x1 + x2)/(n1 + n2)

pc = (44 + 30)/(80 + 100) = 0.41

1 - pc = 1 - 0.41 = 0.59

z = (p1 - p2)/√pc(1 - pc)(1/n1 + 1/n2)

z = (0.55 - 0.3)/√(0.41)(0.59)(1/80 + 1/100) = 3.39

Test statistic = 3.39

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Ira Lisetskai [31]

Answer:

420

Step-by-step explanation:

The first step is to convert 20% to decimal form. 20%=20/100=0.2. Now, you can multiply this by 350 to get 70. Adding this to the original value of 350, you get 350+70=420. Hope this helps!

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Jenna bought enough lattes for all her friends. Each latte was $3.50
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Cylinder has a height of 15 centimeters and a radius of 10 centimeters. What is its volume? Use ​ ≈ 3.14 and round your answer t
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3 years ago
if you roll a fair 6-sided die 9 times, what is the probability that at least 2 of the rolls come up as a 3 or a 4?
Kay [80]

Using the binomial distribution, it is found that there is a 0.857 = 85.7% probability that at least 2 of the rolls come up as a 3 or a 4.

For each die, there are only two possible outcomes, either a 3 or a 4 is rolled, or it is not. The result of a roll is independent of any other roll, hence, the <em>binomial distribution</em> is used to solve this question.

Binomial probability distribution

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • There are 9 rolls, hence n = 9.
  • Of the six sides, 2 are 3 or 4, hence p = \frac{2}{6} = 0.3333

The desired probability is:

P(X \geq 2) = 1 - P(X < 2)

In which:

P(X < 2) = P(X = 0) + P(X = 1)

Then

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{9,0}.(0.3333)^{0}.(0.6667)^{9} = 0.026

P(X = 1) = C_{9,1}.(0.3333)^{1}.(0.6667)^{8} = 0.117

Then:

P(X < 2) = P(X = 0) + P(X = 1) = 0.026 + 0.117 = 0.143

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.143 = 0.857

0.857 = 85.7% probability that at least 2 of the rolls come up as a 3 or a 4.

For more on the binomial distribution, you can check brainly.com/question/24863377

7 0
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