Question

Callaway is thinking about entering the golf ball market. The company will make a profit if its market share is more than 20%. A market survey indicates that 140 of 624 golf ball purchasers will buy a Callaway golf ball.Is this enough evidence to persuade Callaway to enter the golf ball market? How would you make the decision if you were Callaway management? Would you use hypothesis testing?

Answer 1

Answer:

$z=\frac{0.224 -0.2}{\sqrt{\frac{0.2(1-0.2)}{624}}}=1.499$  

$p_v =P(Z>1.499)=0.0669$  

If we compare the p value obtained and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion is not significantly higher than 0.2 or 20%.  

Step-by-step explanation:

How would you make the decision if you were Callaway management? Would you use hypothesis testing?

The best way to test the claim if with a proportion test. The procedure is explained below.

1) Data given and notation

n=624 represent the random sample taken

X=140 represent the golf ball purchasers will buy a Callaway golf ball

$\hat p=\frac{140}{624}=0.224$ estimated proportion of golf ball purchasers will buy a Callaway golf ball

$p_o=0.2$ is the value that we want to test

$\alpha$ represent the significance level

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 true proportion is more than 0,2 or 20%:  

Null hypothesis:$p\leq 0.2$  

Alternative hypothesis:$p > 0.2$  

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:  

$z=\frac{0.224 -0.2}{\sqrt{\frac{0.2(1-0.2)}{624}}}=1.499$  

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 is not provided but let's assume $\alpha=0.05$. The next step would be calculate the p value for this test.  

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

$p_v =P(Z>1.499)=0.0669$  

If we compare the p value obtained and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion is not significantly higher than 0.2 or 20%.