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bekas [8.4K]
2 years ago
11

One of the authors and some statistician friends have an ongoing series of Euchre games that will stop when one of the two teams

is deemed to be statistically significantly better than the other team. Euchre is a card game and each game results in a win for one team and a loss for the other. Only two teams are competing in this series, which we'l call team 1 and team 2.
Required:
a. Define the parameter of interest.
b. What are the null and alternative hypotheses if the goal is to determine if either team is statistically significantly better than the other at winning Euchre?
Mathematics
1 answer:
RSB [31]2 years ago
7 0

Answer:

Parameter of interest is the proportion of games won by a certain team.

H0 : p = 0.5

H1 : p ≠ 0.5

Step-by-step explanation:

Parameter refers to any statistical measure which is derived from the population data. Therefore, the parameter of interest in the scenario described above is the proportion of win or loss by a particular team. This could be either team A or team B.

Probability of either of either win or loss :

Proportion, p of win = 0.5

Null hypothesis ; H0 : p = 0.5

Alternative hypothesis ; H1 : p ≠ 0.5

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Multiply 40x7 then subtract that total you got from multiplying 40x7 with 420

40x7=160

420-160=260

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4-21. Kelso's mom wants to put a floating blanket over the family's circular wading pool to keep the heat in and the leaves out.
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Answer:

a. 78.6 square feet

b. $7.86

Step-by-step explanation:

a. This simply asks us to calculate the area of the 10 feet diameter pool

Formula for area of circle is ;

pi * r^2 = pi * d^2/4

so Area = 22/7 * 10^2/4 = 78.60 square feet

b. This will be the area of the blanket multiplied by the price per square foot of the blanket

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78.60 * 0.1 = $7.86

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3 years ago
What are factors of 27?
Artyom0805 [142]
The answer would be 3^33
6 0
3 years ago
Read 2 more answers
Find the coefficient of variation for each of the two sets of data, then compare the variation. Round results to one decimal pla
svp [43]

Here is  the correct computation of the question given.

Find the coefficient of variation for each of the two sets of data, then compare the variation. Round results to one decimal place. Listed below are the systolic blood pressures (in mm Hg) for a sample of men aged 20-29 and for a sample of men aged 60-69.

Men aged 20-29:      117      122     129      118     131      123

Men aged 60-69:      130     153      141      125    164     139

Group of answer choices

a)

Men aged 20-29: 4.8%

Men aged 60-69: 10.6%

There is substantially more variation in blood pressures of the men aged 60-69.

b)

Men aged 20-29: 4.4%

Men aged 60-69: 8.3%

There is substantially more variation in blood pressures of the men aged 60-69.

c)

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

d)

Men aged 20-29: 7.6%

Men aged 60-69: 4.7%

There is more variation in blood pressures of the men aged 20-29.

Answer:

(c)

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

Step-by-step explanation:

From the given question:

The coefficient of variation can be determined by the relation:

coefficient \ of  \ variation = \dfrac{standard \ deviation}{mean}*100

We will need to determine the coefficient of variation both men age 20 - 29 and men age 60 -69

To start with;

The coefficient of men age 20 -29

Let's first find the mean and standard deviation before we can do that ;

SO .

Mean = \dfrac{\sum \limits^{n}_{i-1}x_i}{n}

Mean = \frac{117+122+129+118+131+123}{6}

Mean = \dfrac{740}{6}

Mean = 123.33

Standard deviation  = \sqrt{\dfrac{\sum (x_i- \bar x)^2}{(n-1)} }

Standard deviation =\sqrt{\dfrac{(117-123.33)^2+(122-123.33)^2+...+(123-123.33)^2}{(6-1)} }

Standard deviation  = \sqrt{\dfrac{161.3334}{5}}

Standard deviation = \sqrt{32.2667}

Standard deviation = 5.68

The coefficient \ of  \ variation = \dfrac{standard \ deviation}{mean}*100

coefficient \ of  \ variation = \dfrac{5.68}{123.33}*100

Coefficient of variation = 4.6% for men age 20 -29

For men age 60-69 now;

Mean = \dfrac{\sum \limits^{n}_{i-1}x_i}{n}

Mean = \frac{   130 +    153    +  141  +    125 +   164  +   139}{6}

Mean = \dfrac{852}{6}

Mean = 142

Standard deviation  = \sqrt{\dfrac{\sum (x_i- \bar x)^2}{(n-1)} }

Standard deviation =\sqrt{\dfrac{(130-142)^2+(153-142)^2+...+(139-142)^2}{(6-1)} }

Standard deviation  = \sqrt{\dfrac{1048}{5}}

Standard deviation = \sqrt{209.6}

Standard deviation = 14.48

The coefficient \ of  \ variation = \dfrac{standard \ deviation}{mean}*100

coefficient \ of  \ variation = \dfrac{14.48}{142}*100

Coefficient of variation = 10.2% for men age 60 - 69

Thus; Option C is correct.

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

4 0
3 years ago
Josie found a small bird bath at a garage sale. The bird bath has a circular opening with a radius of 10 cm, as shown in the dia
sveticcg [70]

Answer: 1256cm

Step-by-step explanation:

8 0
3 years ago
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