Betting the Spreads. 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 uniforn betting across teams, Las Vegas oddsmakers assign a point spread to each game. The point spread is the oddsmakers" prediction for th number of points by which the favored team will win. If you bet of 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 $1-$Seed?" (Chance, Vol. 11(1), pp. $27-31$). 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 spread 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 answer in part (b).

Step by step solution

01

Part (a) Given information

With a standard deviation of $10.9$ points, the mean point-spread error was found to be $0.2$ point. A point-spread inaccuracy of $0$ means that the point spread was a flawless estimation of the two teams' relative talents for that particular game.
a. What is the (population) mean point-spread error if the oddsmakers estimate properly on average?

02

Explanation

Because the perfect estimate is $0$ points, the population mean point-spread error is $0$ points. As a result, the ideal estimate of the mean point spread error is $0$ points.

03

Part(b) Step 1: To find

By the $5\%$significance level, the mean point spread error.

04

Given information

With a standard deviation of $10.9$ points, the mean point-spread error was found to be $0.2$ point. A point-spread inaccuracy of $0$ means that the point spread was a flawless estimation of the two teams' relative talents for that particular game.
b. Using the data, determine whether the population mean point spread error differs from $0$at the $5\%$significance level.

05

Calculation

The t test will be used first at the $5\%$significance level (since the sample is quite large and the distribution is approximately normal):

$H_0:\mu =0,H_a:\mu \ne 0$

Find the $t$-value now:

$$\begin{gathered} t=\frac{-0.2-0}{10.9/\sqrt{2109}} \\ \approx -0.84 \end{gathered}$$

Using table $IV$ , find the corresponding $P$- value:

$$\begin{gathered} P>2\times 0.10 \\ =0.20 \end{gathered}$$

Reject the null hypothesis if the P-value is less than or equal to the significance level:

$$\begin{gathered} P>0.05 \\ =5\% \end{gathered}$$,
Do not reject width="21" height="24" role="math">H0

Then, the solution is $P>0.05=5\%$
$H_o$should not be rejected.

06

Part (c) Step 1: Given information

To locate the part's interpretation (b).

07

Explanation

Given that,
With a standard deviation of $10.9$ points, the mean point-spread error was found to be$0.2$ point. A point-spread inaccuracy of $0$means that the point spread was a flawless estimation of the two teams' relative talents for that particular game.
c. Partially interpret your response (b).
Calculation:
The data are insufficient to determine that the population mean point-spread error is greater than zero.

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