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? They obtained the difference between the actual margin of victory and the point spread, called the point-spread error for \(2109\) college basketball games. The mean-point spread error was found to be \(-0.2\) point with a standard deviation of \(10.9\) points. For a particular game, a point spread error of \(0\) indicates that the point was a perfect estimate of the two teams relative abilities.

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

b. Use the data to decide, at the \(5%\) significance level, whether the (population) mean point-spread error differs from \(0\).

c. Interpret your answers in part (b).

Short answer

Part a. \(0\) points.

Part b. \(P>0.05=5%\Rightarrow Do not reject H_{0}\).

Part c. The data do not provide sufficient evidence to conclude that the population mean point-spread error differs from \(0\)

Step-by-step solution

Part a. Step 1. Given information

The mean point- spread error was found to be \(-0.2\) point with a standard deviation of \(10.9\) points. For a particular game, a point-spread error of \(0\) indicates that the point spread was a perfect estimate of the two teams’ relative abilities.

If, on average, the odds makers are estimating correctly, what is the (population) mean point- spread error?

Part a. Step 2. Calculation

The population mean point-spread error consists \(0\) points, because the perfect estimate is \(0\) points.

Thus, the perfect estimation for the mean point spread error is \(0\) points.

Part b. Step 1. Given information

The mean point- spread error was found to be \(-0.2\) point with a standard deviation of \(10.9\) points. For a particular game, a point-spread error of \(0\) indicates that the point spread was a perfect estimate of the two teams’ relative abilities.

Use the data to decide, at the \(5%\) significance level, whether the population mean point spread error differ from \(0\).

Part b. Step 2. Calculation

At the \(5%\) significance level, firstly will use the \(t-\)test (since the sample is very large and thus the distribution is approximately normal):

\(H_{0}:\mu =0\)

\(H_{a}:\mu \neq0\)

Now find the \(t-\)value:

\(t=\frac{-0.2-0}{10.9 / \sqrt{2109}}\approx -0.84\)

Find the corresponding \(P-\)value using table IV:

\(P>2\times 0.10=0.20\)

If the \(P-\)value is smaller than or equal to the significance level, reject the null hypothesis:

\(P>0.05=5%\Rightarrow\) do not reject \(H_{0}\)

Thus, the solution is \(P>0.05=5%\Rightarrow\) do not reject \(H_{0}\) .

Part c. Step 1. Explanation

The data do not provide sufficient evidence to conclude that the population mean point-spread error differs from \(0\).