Question

Vermont-based Green Mountain Coffee Roasters dominates the market for single-serve coffee in the United States, with its subsidiary Keurig accounting for approximately 70% of sales ("Rivals Try to Loosen Keurig's Grip on Single-Serve Coffee Market," Chicago Tribune, February 26, 2011). But Keurig's patent on K-cups, the plastic pods used to brew the coffee, is expected to expire in 2012, allowing other companies to better compete. Suppose a potential competitor has been conducting blind taste tests on its blend and finds that 47% of consumers strongly prefer its French Roast to that of Green Mountain Coffee Roasters. After tweaking its recipe, the competitor conducts a test with 144 tasters, of which 72 prefer its blend. The competitor claims that its new blend is preferred by more than 47% of consumers to Green Mountain Coffee Roasters' French Roast.

Refer to Exhibit 9-7. At the 1% significance level, does the evidence support the claim?

a. No, since the value of the test statistic is less than the critical value
b. Yes, since the value of the test statistic is less than the critical value
c. No, since the value of the test statistic is greater than the critical value
d. Yes, since the value of the test statistic is greater than the critical value

Answer 1

Answer:

a. No, since the value of the test statistic is less than the critical value

Step-by-step explanation:

1) Data given and notation

n=144 represent the random sample taken

X=72 represent the number of people that prefer the blend

$\hat p=\frac{72}{144}=0.5$ estimated proportion of people that prefer the blend

$p_o=0.47$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.959

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 if higher than 0.47:  

Null hypothesis:$p\leq 0.47$  

Alternative hypothesis:$p > 0.47$  

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:  

$z=\frac{0.5 -0.47}{\sqrt{\frac{0.47(1-0.47)}{144}}}=0.721$  

4) Statistical decision  

We can calculate the critical value since we have a right tailed test, we need to look into the normal standard distribution a value that accumulates 0.01 of the area on the right and 0.99 on the left. And this value is:

$z_{\alpha/2}=2.33$

And we can use the following excel code to find the critical value: "=NORM.INV(0.99,0,1)"

Our calculated value on this case is less than the critical value so the best conclusion is:

a. No, since the value of the test statistic is less than the critical value