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harina [27]
3 years ago
13

A social scientist is interested in determining if there is a significant difference in the proportion of Republicans between tw

o areas of town. He takes independent random samples of 200 families in each area of town and a significance test was conducted. The p-value was 0.0156. What should be our conclusions?
Mathematics
2 answers:
hodyreva [135]3 years ago
4 0

Answer:

Given

p-value = 0.0156

n = 200 families

Our conclusion goes as follows:

The statistical evidence is pretty strong enough to conclude that there is a difference in the proportion of Republicans between the two areas of town because we know that small p­-value indicates significant differences, and a p­value of 0.0156 is pretty small to ascertain our conclusion.

Shtirlitz [24]3 years ago
3 0

Answer:

Given a p-value of 0.0156

n= 200

Since the p-value(0.0156) is less than 0.05, we can conclude that there is a difference in the proportion of republicans in the two areas.

A small p-value indicates significant difference and a p-value that is less than or equals (≤) 0.05 is small. In this case, the p-value 0.0156 is quite small. Therefore, we can say there is a significant difference.

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7 0
3 years ago
Suppose n people, n ≥ 3, play "odd person out" to decide who will buy the next round of refreshments. The n people each flip a f
blondinia [14]

Answer:

Assume that all the coins involved here are fair coins.

a) Probability of finding the "odd" person in one round: \displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1}.

b) Probability of finding the "odd" person in the kth round: \displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1} \cdot \left( 1 - n \cdot \left(\frac{1}{2}\right)^{n - 1}\right)^{k - 1}.

c) Expected number of rounds: \displaystyle \frac{2^{n - 1}}{n}.

Step-by-step explanation:

<h3>a)</h3>

To decide the "odd" person, either of the following must happen:

  • There are (n - 1) heads and 1 tail, or
  • There are 1 head and (n - 1) tails.

Assume that the coins here all are all fair. In other words, each has a 50\,\% chance of landing on the head and a

The binomial distribution can model the outcome of n coin-tosses. The chance of getting x heads out of

  • The chance of getting (n - 1) heads (and consequently, 1 tail) would be \displaystyle {n \choose n - 1}\cdot \left(\frac{1}{2}\right)^{n - 1} \cdot \left(\frac{1}{2}\right)^{n - (n - 1)} = n\cdot \left(\frac{1}{2}\right)^n.
  • The chance of getting 1 heads (and consequently, (n - 1) tails) would be \displaystyle {n \choose 1}\cdot \left(\frac{1}{2}\right)^{1} \cdot \left(\frac{1}{2}\right)^{n - 1} = n\cdot \left(\frac{1}{2}\right)^n.

These two events are mutually-exclusive. \displaystyle n\cdot \left(\frac{1}{2}\right)^n + n\cdot \left(\frac{1}{2}\right)^n  = 2\,n \cdot \left(\frac{1}{2}\right)^n = n \cdot \left(\frac{1}{2}\right)^{n - 1} would be the chance that either of them will occur. That's the same as the chance of determining the "odd" person in one round.

<h3>b)</h3>

Since the coins here are all fair, the chance of determining the "odd" person would be \displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1} in all rounds.

When the chance p of getting a success in each round is the same, the geometric distribution would give the probability of getting the first success (that is, to find the "odd" person) in the kth round: (1 - p)^{k - 1} \cdot p. That's the same as the probability of getting one success after (k - 1) unsuccessful attempts.

In this case, \displaystyle p = n \cdot \left(\frac{1}{2}\right)^{n - 1}. Therefore, the probability of succeeding on round k round would be

\displaystyle \underbrace{\left(1 - n \cdot \left(\frac{1}{2}\right)^{n - 1}\right)^{k - 1}}_{(1 - p)^{k - 1}} \cdot \underbrace{n \cdot \left(\frac{1}{2}\right)^{n - 1}}_{p}.

<h3>c)</h3>

Let p is the chance of success on each round in a geometric distribution. The expected value of that distribution would be \displaystyle \frac{1}{p}.

In this case, since \displaystyle p = n \cdot \left(\frac{1}{2}\right)^{n - 1}, the expected value would be \displaystyle \frac{1}{p} = \frac{1}{\displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1}}= \frac{2^{n - 1}}{n}.

7 0
3 years ago
Simplify the following expresion by combining like terms 6y-23-y+72
Maurinko [17]
Simplest form would be, 5y-23+72 then it would be 5y+49
7 0
3 years ago
PLEASE HELP!!!!! 1-Suppose the population of a city is 50000 and growing 3% each year.
Hunter-Best [27]

Answer:

A 104,689

Step-by-step explanation:

y= a x b^x

y=50,000 x 1.03^x

7 0
3 years ago
club a raised $168 by washing 42 cars. club b raised $152 by washing 38 cars. are these fundraising rates equivalent ? explain y
uysha [10]

Answer:

The two rates are equivalent.

Step-by-step explanation:

Here we'll find the respective unit rates and then compare those rates:

$168

----------- = $4/car

42 cars

$152

----------- = $4/car

38 cars

These two rates are precisely the same.  Thus, we conclude that they are equivalent.

6 0
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