Question

College basketball, and particularly the NCAA basketball tournament is a popular venue for gambling, from novices in office betting pools to high rollers. To encourage uniform betting across teams, Las Vegas oddsmakers assign a point spread to each game. The point spread is the oddsmakers' prediction for the number of points by which the favored team will win. If you bet on the favorite, you win the bet provided the favorite wins by more than the point spread; otherwise, you lose the bet. Is the point spread a good measure of the relative ability of the two teams? H. Stern and B. Mock addressed this question in the paper "College Basketball Upsets: Will a 16-Seed Ever Beat a 1-Speed? They obtained the difference between the actual margin of victory and the point spread, called the point-spread error, for 2109college basketball games. The mean point-spread error was found to be -0.2point with a standard deviation of 10.9points. For a particular game, a point-spread error of 0indicates that the point spread was a perfect estimate of the two teams' relative abilities.

Part (a): If, on average, the oddsmakers are estimating correctly, what is the (population) mean point-spread error?

Part (b): Use the data to decide, at the 5%significance level, whether the (population) mean point-spread error?

Part (c): Interpret your answer in part (b).

Short answer

Part (a): The mean point-spread error is 0.

Part (b): The test results are not statistically significant at 5% level of significance.

Part (c): There is no sufficient evidence to conclude that the mean point-spread error differ from 0.

Step-by-step solution

Part (a) Step 1. Given information.

Consider the given question,

The mean point-spread error is -0.2.

The standard deviation is 10.9.

The significance level is $\alpha =0.05$.

Part (a) Step 2. Determine the mean point-spread error.

Consider the given question,

The mean point-spread error is 0 if the oddsmakers are estimating correctly.

Part (b) Step 1. Check whether or not the mean point-spread error differs from 0.

The null hypothesis is given below,

$H_0:\mu =0$

The mean point-spread error is0.

The alternative hypothesis is given below,

$H_a:\mu \ne 0$

The mean point-spread error differs from 0.

Part (b) Step 2. Compute the value of the test statistics and P-value.

On computing the value of the test statistics and P-value,

1. Choose Stat>Basic Statistics>1-Sample t.
2. In Summarized data, enter the sample size 2109 and mean -0.2.
3. In Standard deviation, enter 10.9.
4. In Perform hypothesis test, enter the test mean as 0.
5. Check Options, enter Confidence level as 95.
6. Choose not equal in alternative.
7. Click OK in all dialogue boxes.

Hence, from the MINITAB output, the value of test statistics is -0.84 and the P-value is 0.4.

Part (b) Step 3. Write the conclusion.

If $P\le \alpha$, then reject the null hypothesis.

The P-value is 0.4which is greater than the level of significance that is $P\left(=0.4\right)>\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is not rejected at 5% level.

It can be concluded that the test results are not statistically significant at 5% level of significance.

Part (c) Step 1. Write the interpretation.

Consider the part (b),

On interpreting, we can say that there is no sufficient evidence to conclude that the mean point-spread error differ from0.