7.4.1.3 - Example: Proportion NFL Coin Toss Wins

Research question: Is the proportion of NFL overtime coin tosses that are won different from 0.50?

StatKey was used to construct a randomization distribution:

175 200 150 100 0.42 0.44 0.46 0.48 0.50 null = 0.5 0.52 0.54 0.56 0.58 125 75 50 25 0 Randomization Test for a Proportion Original Sample Count 240 428 0.561 186 428 0.435 Sample Size Proportion Count Sample Size Proportion Randomization Sample Randomization Dotplot of Null hypothesis: p = NFL Coin Flip Wins Overtime Generate 1 Sample Generate 10 Samples Generate 100 Samples Generate 1000 Samples Proportion 0.5 Reset Plot Edit Data Left Tail Two - Tail Right Tail

 

Step 1: Check assumptions and write hypotheses

From the given StatKey output, the randomization distribution is approximately normal.

\(H_0\colon p=0.50\)

\(H_a\colon p \ne 0.50\)

Step 2: Calculate the test statistic

\(test\;statistic=\dfrac{sample\;statistic-null\;parameter}{standard\;error}\)

The sample statistic is the proportion in the original sample, 0.561. The null parameter is 0.50. And, the standard error is 0.024.

\(test\;statistic=\dfrac{0.561-0.50}{0.024}=\dfrac{0.061}{0.024}=2.542\)

Step 3: Determine the p value

The p value will be the area on the z distribution that is more extreme than the test statistic of 2.542, in the direction of the alternative hypothesis. This is a two-tailed test:

Distribution Plot Normal, Mean=0, StDev=1 0.0 0.1 0.2 0.3 0.4 0 X Density -2.54200 0.0055110 0.0055110 2.542

The p value is the area in the left and right tails combined: \(p=0.0055110+0.0055110=0.011022\)

Step 4: Make a decision

The p value (0.011022) is less than the standard 0.05 alpha level, therefore we reject the null hypothesis.

Step 5: State a "real world" conclusion

There is evidence that the proportion of all NFL overtime coin tosses that are won is different from 0.50