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larisa [96]
2 years ago
6

According to a soccer coach, 75% of soccer players have had at least one sprained ankle. An athletic trainer would like to inves

tigate this claim. To do so, the trainer selects a random sample of 125 college soccer players from across the country and finds that 99 of them have had at least one sprained ankle. The trainer would like to know if the data provide convincing evidence that the true proportion of college soccer players who have had at least one sprained ankle is greater than 75%. What are the appropriate hypotheses for this test
Mathematics
1 answer:
Greeley [361]2 years ago
5 0

Answer:

H0 : μ = 0.75

H1 : μ > 0.75

Step-by-step explanation:

Given :

Sample size, n = 125

x = 99

Phat = x / n = 99 / 125 = 0.792

Population proportion, P = 0.75

The hypothesis :

Null hypothesis :

H0 : μ = 0.75

Alternative hypothesis ;

Egates the null hypothesis ; since the sample proportion is greater than the the population proportion or claim "; then we use the greater than sign.

H1 : μ > 0.75

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miss Akunina [59]

Answer:

The probability that the intersection will come under the emergency program is 0.1587.

Step-by-step explanation:

Lets divide the problem in months rather than in years, because it is more suitable to divide the period to make a better approximation. If there were 36 accidents in average per year, then there should be 3 accidents per month in average. We can give for the amount of accidents each month a Possion distribution with mean 3 and variance 3.

Since we want to observe what happen in a period of one year, we will use a sample of 12 months and we will take its mean. We need, in average, more than 45/12 = 3.75 accidents per month to confirm that the intersection will come under the emergency program.

For the central Limit theorem, the sample mean will have a distribution Normal with mean 3 and variance 3/12 = 0.25; thus its standard deviation is √0.25 = 1/2.

Lets call the sample mean distribution X. We can standarize X obtaining a standard Normal random variable W with distribution N(0,1).

W = \frac{X-\mu}{\sigma} = \frac{X-3}{1/2} = 2x-6

The values of \phi , the cummulative distribution function of W, can be found in the attached file. We are now ready to compute the probability of X being greater than 3.75, or equivalently, the probability than in a given year the amount of accidents is greater than 45, leading the intersection into an emergency program

P(X > 3.75) = P(2X-6 > 2*3.75-6) = P(W > 1) = 1-\phi(1) = 1-0.8413 \\= 0.1587

Download pdf
7 0
3 years ago
Ania scored a total of 120 points this season so far and averaged 15 points a game. How many games has she played?​
arlik [135]
She played 8 games. Since she earned a total of 120 points and 15 points per game you would divide 120/15=8
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2 years ago
Im on a girls basketball team , 3 are in sixth grade , 5 are in seventh grade , and 6 are in eighth grade . so which is the rati
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Answer:

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3 years ago
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B.
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