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

Answer:

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

$p_v =P(Z>2.958)=0.0015$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ 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 .  

Step-by-step 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:  

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (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:  

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

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 $p_v$ 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 .