any help is appreciated

Question

Let X be the number of goal chances that result in ' goals for a football team during n goal chances. In the course of many matches, the team scores on 20% of the chances.

a)

Explain why it might be reasonable (at least as an approximation) to assume that X is

binomially distributed in this situation. Do this by going through each of the points that

must be fulfilled for us to be able to use binomial distribution, consider whether it is reasonable that

they are met here and specify any additional assumptions we must make.

What is the probability that the team scores two goals in a match with n = 12 goal chances?

What is the probability that they score more than 50 goals during a season if they score

n = 300 goal chances?

Before a new season, the team acquires striker Tony to increase the probability of scoring on goal chances. During the first 116 scoring chances, they scored 29 goals.

b)

Determine with a hypothesis test whether the probability of scoring on the goal chances has increased. use 5%

level. Also calculate the test's p-value.

What would the conclusion of the test be if we had used the 10% level?

Answer 1

A) Let's look at the conditions of the binomial distribution:

1) Is n fixed? That appears to be the case since we define X as the number of goals in n chances.

2) Are all goal chances independent? Momentum is definitely a force in games, although it is somewhat difficult to quantify. Because of this, you can assume all of the goal chances are independent, at least for approximation purposes.

3) Do only 2 outcomes exist? Yes. The team either scores a goal or doesn't score.

Now that we've established that for approximation purposes, X is a binomially distributed variable, we know that the p-value can be calculated by the formula:

P(X=r) = _nC_r•0.2^r•0.8^n-r

This means that for the team to score 2 goals on 12 goal chances, that probability is

P(X=2) = ₁₂C₂•0.2²•0.8¹⁰ = 66(.04)(0.1073741824) ≈ 0.2835

There is a 28.35% chance of the team scoring 2 goals in 12 goal chances.

Now for scoring more than 50 goals in 300 chances, normal computations would take forever unless you have a computer doing that work for you. Thus, we would like to approximate this probability with a normal distribution. We can do this because np = 60 > 5 and n(1-p) = 240 > 5. This normal distribution would have the following parameters:

μ = np = 300(0.2) =5 60

σ = √(np(1-p)) = √(300•0.2•0.8) ≈ 6.9282

Then to find the probability of getting more than 50 goals, we find a z-score and the corresponding probability on a chart (or in a calculator if available). NOTE: because we are using a normal distribution (continuous distribution) to approximate a binomial distribution (discrete distribution), the point of interest will be 49.5 as (1) all values between 49.5 and 50.5 will round to 50 and (2) we're looking for "more than" probability, so the lower side is used.

z = (X-μ)/σ = (49.5-60)/6.9282) = -1.52

P(z>-1.52) = 0.9357

Thus there is a 93.57% chance of the team scoring more than 50 goals in 300 chances.

B) For this hypothesis test, we'll use the normal distribution above. The assumed mean (from the null hypothesis) is 20% of 116, giving us μ=23.2; the standard deviation also changes because of n=116 to σ=4.308. This makes the test statistic and corresponding probability equal to:

z* = (28.5 - 23.2)/4.308 = 1.23

P(z>0.765) = 0.1093. **Don't multiply this by 2; "increase" means this is a one-tail test**

Since the p-value is greater than 5% (or even 10% for that matter), we don't have enough evidence to reject the null hypothesis. Thus, there is insufficient evidence to conclude an increase in the probability of a team scoring on a scoring chance.