Question

LeRoy, a starting player for a major college basketball team, made only 40% of his free throws last season. During the summer, he worked on developing a softer shot in hopes of improving his free throw accuracy. In the first eight games of this season, LeRoy made 25 free throws in 40 attempts. You want to investigate whether LeRoy’s work over the summer will result in a higher proportion of free-throw successes this season. What conclusion would you draw at the α = 0.01 level about LeRoy’s free throw shooting?

Answer 1

Answer:

$z=\frac{0.625 -0.4}{\sqrt{\frac{0.4(1-0.4)}{40}}}=2.905$  

$p_v =P(z>2.905)=0.0018$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of free throws made is higher than 0.4

Step-by-step explanation:

Data given and notation

n=40 represent the random sample taken

X=25 represent the number of free throws made

$\hat p=\frac{25}{40}=0.625$ estimated proportion of free throws made

$p_o=0.4$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

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

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.4:  

Null hypothesis:$p\leq 0.4$  

Alternative hypothesis:$p > 0.4$  

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 $p_o$.

Calculate the statistic  

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

$z=\frac{0.625 -0.4}{\sqrt{\frac{0.4(1-0.4)}{40}}}=2.905$  

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 $\alpha=0.01$. The next step would be calculate the p value for this test.  

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

$p_v =P(z>2.905)=0.0018$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of free throws made is higher than 0.4