Question

Coffee: A popular chain of cafes has been receiving online complaints about one store location. Regular customers complained that the staff at this location consistently underfill their cups of coffee. The owner of the chain will personally visit the location to meet with the staff to determine if more than 10% of the 16 ounce cups are underfilled by 1 fluid ounce. She arranges to have "mystery shoppers" visit the store. Mystery shoppers are people who are paid to pose as a regular customer and provide feedback to the owner about customer service. After 100 mystery shop visits, she finds that 18 of the 16 ounce cups of coffee were underfilled. She plans to test the hypotheses H 0 : p = 0.10 versus H a : p > 0.10 .
What is the test statistic?


z = -2.67

z = 2.67

z = 2.08

z = -2.08

Answer 1

Answer:

$z=\frac{0.18 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=2.67$  

Step-by-step explanation:

1) Data given and notation

n=100 represent the random sample taken

X=18 represent the ounce cups of coffee that were underfilled

$\hat p=\frac{18}{100}=0.18$ estimated proportion of ounce cups of coffee that were underfilled

$p_o=0.1$ 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 it's higher than 0.1 or 10%:  

Null hypothesis:$p\leq 0.1$  

Alternative hypothesis:$p > 0.1$  

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:  

$z=\frac{0.18 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=2.67$  

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 assumed for this case is $\alpha=0.05$. The next step would be calculate the p value for this test.  

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

$p_v =P(Z>2.67)=0.0037$  

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 thetrue proportion is not significanlty higher than 0.1 or 10% .