Question

There were 2430 Major League Baseball games played in 2009, and the home team won the game in 53% of the games. If we consider the games played in 2009 as a sample of all MLB games, test to see if there is evidence, at the 5% level, that the home team wins more than half the games. (Show all steps of the hypothesis test to receive points)

Answer 1

`z=\frac{0.53 -0.5}{\sqrt{(0.5(1-0.5))/(2430)}}=2.958`

`p_v =P(Z>2.958)=0.0015`

So the p value obtained was a very low value and using the significance level given
`\alpha=0.05` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion is significantly higher than 0.5 .

Explanation:

1) Data given and notation

n=2430 represent the random sample taken

`\hat p=0.53` estimated proportion when the home team won the game

`p_o=0.5` is the value that we want to test

`\alpha=0.05` represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

`p_v` represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the proportion is higher than 0.5 or 50%:

Null hypothesis:
`p\leq 0.5`

Alternative hypothesis:
`p > 0.5`

When we conduct a proportion test we need to use the z statistic, and the is given by:

`z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}}` (1)

The One-Sample Proportion Test is used to assess whether a population proportion
`\hat p` is significantly different from a hypothesized value
`p_o`.

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

`z=\frac{0.53 -0.5}{\sqrt{(0.5(1-0.5))/(2430)}}=2.958`

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided
`\alpha=0.05`. The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

`p_v =P(Z>2.958)=0.0015`

So the p value obtained was a very low value and using the significance level given
`\alpha=0.05` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion is significantly higher than 0.5 .