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hodyreva [135]
3 years ago
10

In the local frisbee league, teams have 7 members and each of the 4 teams takes turns hosting tournaments. At each tournament, e

ach team selects two members of that team to be on the tournament committee, except the host team, which selects three members. How many possible 9 member tournament committees are there?
Mathematics
1 answer:
mars1129 [50]3 years ago
8 0

There are 4 teams in total and each team has 7 members. One of the team will be the host team.

Tournament committee will be made from 3 members from the host team and 2 members from each of the three remaining teams. Selecting the members for tournament committee is a combinations problem. We have to select 3 members out 7 for host team and 2 members out of 7 from each of the remaining 3 teams.

So total number of possible 9 member tournament committees will be equal to:

7C3 \times 7C2 \times 7C2 \times 7C2\\ \\  = 324135

This is the case when a host team is fixed. Since any team can be the host team, there are 4 possible ways to select a host team. So the total number of possible 9 member tournament committee will be:

9 \times 324135\\ \\  =2917215

Therefore, there are 2917215 possible 9 member tournament committees

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Answer:

We conclude that there is a difference in seat belt use.

Step-by-step explanation:

We are given that of 1,000 drivers 16-24 years old, 79% said they buckle up, whereas 924 of 1,100 drivers 25-69 years old said they did.

<u><em>Let </em></u>p_1<u><em> = population proportion of drivers 16-24 years old who buckle up .</em></u>

<u><em /></u>p_2<u><em> = population proportion of drivers 25-69 years old who buckle up .</em></u>

So, Null Hypothesis, H_0 : p_1-p_2 = 0      {means that there is no significant difference in seat belt use}

Alternate Hypothesis, H_A : p_1-p_2\neq 0      {means that there is a difference in seat belt use}

The test statistics that would be used here <u>Two-sample z proportion statistics;</u>

                     T.S. =  \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }  ~ N(0,1)

where, \hat p_1 = sample proportion of drivers 16-24 years old who buckle up = 79%

\hat p_2 = sample proportion of drivers 25-69 years old who buckle up = \frac{924}{1100} = 84%

n_1 = sample of 16-24 years old drivers = 1000

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So, <u><em>test statistics</em></u>  =  \frac{(0.79-0.84)-(0)}{\sqrt{\frac{0.79(1-0.79)}{1000}+\frac{0.84(1-0.84)}{1100} } }  

                              =  -2.946

The value of z test statistics is -2.946.

Since, in the question we are not given with the level of significance so we assume it to be 5%. <u><em>Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.</em></u><em> </em>

<em>Since our test statistics doesn't lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><em><u>we reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that there is a difference in seat belt use.

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