Marcus believes he can score a goal at least 70% of the time.

He runs a simulation where he tracks the number of successful goals out of ten attempts. He runs the simulation 100 times.

Goals 0 1 2 3 4 5 6 7 8 9 10
Frequency 0 1 2 2 1 0 1 18 40 30 5

He decides to use a significance level of 0.05. What can he conclude about his null hypothesis that he will score no more than 6 goals out of 10?

A) The p-value is close to 0. Fail to reject the null hypothesis.
B) The p-value is close to 1. Fail to reject the null hypothesis.
C) The p-value is close to 0. Reject the null hypothesis.
D) The p-value is close to 1. Reject the null hypothesis.

1 answer

To analyze the simulation results and determine whether Marcus can reject the null hypothesis that he will score no more than 6 goals out of 10 attempts, we should first summarize the information provided in the frequency distribution of successful goals scored:

  • Total simulations = 100
  • Goals scored (successful attempts) ranging from 0 to 10, with the corresponding frequencies as follows:
    • 0 goals: 0 times
    • 1 goal: 1 time
    • 2 goals: 2 times
    • 3 goals: 2 times
    • 4 goals: 1 time
    • 5 goals: 0 times
    • 6 goals: 1 time
    • 7 goals: 18 times
    • 8 goals: 40 times
    • 9 goals: 30 times
    • 10 goals: 5 times

Now, we need to calculate how many times Marcus scored more than 6 goals. Those frequencies are:

  • 7 goals: 18 times
  • 8 goals: 40 times
  • 9 goals: 30 times
  • 10 goals: 5 times

Adding those:

  • Total number of times scored more than 6 goals = 18 + 40 + 30 + 5 = 93.

To calculate the p-value, we first determine the proportion of times he scored more than 6 goals:

\[ \text{Proportion} = \frac{93}{100} = 0.93. \]

Since the null hypothesis posited that he scores no more than 6 goals, we need to find the probability (p-value) of scoring more than 6 goals. The obtained proportion of 0.93 indicates that there is a very high likelihood (93%) of scoring more than 6 goals.

Next, we compare this to the significance level of 0.05:

  1. If the p-value is close to 0, we would reject the null hypothesis.
  2. If the p-value is close to 1, we would fail to reject the null hypothesis.

Since we found a high proportion (0.93), it suggests that the p-value is indeed close to 0 (very strong evidence against the null hypothesis of scoring no more than 6 goals).

Thus, Marcus can conclude that he has sufficient evidence to reject the null hypothesis.

The correct answer is: C) The p-value is close to 0. Reject the null hypothesis.