PLEASE ANSWER ASAP + BRAINLIEST!!!!

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 t

Ivahew [28]

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 .

8 0

3 years ago

Please help me with this!

salantis [7]

The original area of a face would be a^2. Now that you added b to the edge, the new area of each face would be (a+b)^2. To find how much the are increased, subtract a^2 from (a+b)^2. $(a+b)^2-(a^2)= a^2+2ab+b^2-(a^2)= 2ab+b^2= b(2a+b)$ So the answer is b(2a+b)