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katen-ka-za [31]
3 years ago
9

Researchers are studying rates of homeowners in a certain town. They believe that the proportion of people ages 36-50 who own ho

mes is signifificantly greater than the proportin of people age 21-35 who own homes and want to test this claim. The results of the surverys are: Homeowners Renters Total
Ages 21-35 18 38 56



Ages36-50 40 22 62


TOTAL 58 60 118



What are the null hypothesis and alternative hypothesis for this situation
Mathematics
1 answer:
Ira Lisetskai [31]3 years ago
6 0

Answer:

Being p1 the proportion for people of ages 36-50 and p2 the proportion for people of ages 21-35, the null and alternative hypothesis will be:

H_0: p_1-p_2=0\\\\H_a: p_1-p_2>0

Step-by-step explanation:

A hypothesis test on the difference of proportions needs to be performed for this case.

We have two sample proportions and we want to test if the true population proportions differ from each other, usign the information given by the sample statistics.

The claim is that the proportion of people of ages 36-50 who own homes is significantly greater than the proportin of people age 21-35 who own homes.

The term "higher" will define the alternative hypothesis, that is the hypothesis that represents what is claimed. The null hypothesis always include the equal sign, and will state that both proportions do not differ.

Being p1 the proportion for people of ages 36-50 and p2 the proportion for people of ages 21-35, the null and alternative hypothesis will be:

H_0: p_1-p_2=0\\\\H_a: p_1-p_2>0

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Answer:  15

Step-by-step explanation:

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For (a+b)^6 , n= 6

T_3=T_{2+1}=\ ^6C_2(a)^{6-2}(b)^2\\\\

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Hence, the coefficient of the third term in the binomial expansion of  (a+b)^6 is 15.

3 0
3 years ago
Read 2 more answers
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?
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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.
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  • 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

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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
3 years ago
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