You run a competitive K-cup coffee company. Your most popular flavor is the French Roast. Suppose a potential competitor has been conducting blind taste tests on its blend and finds that 48% of consumers strongly prefer its French Roast to yours. 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 48% of consumers to your French Roast. What is the value of the appropriate test statistic to test this claim

We are given that a potential competitor has been conducting blind taste tests on its blend and finds that 48% of consumers strongly prefer its French Roast to yours.

After tweaking its recipe, the competitor conducts a test with 144 tasters, of which 72 prefer its blend.

Let p = proportion of consumers who prefers new blend.

So, Null Hypothesis, [tex]H_0[/tex] : p [tex]\leq[/tex] 48% {means that the new blend is preferred by less than or equal to 48% of consumers}

Alternate Hypothesis, [tex]H_A[/tex] : p > 48% {means that the new blend is preferred by more than 48% of consumers}

The test statistics that would be used here One-sample z proportion statistics;

T.S. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p (1-\hat p)}{n} } }[/tex] ~ N(0,1)

where, [tex]\hat p[/tex] = sample proportion of tasters prefer new blend = [tex]\frac{72}{144}[/tex] = 0.50

n = sample of tasters = 144

So, test statistics = [tex]\frac{0.50-0.48}{\sqrt{\frac{0.50 (1-0.50)}{144} } }[/tex]

= 0.48

Hence, the value of the appropriate z test statistic to test this claim is 0.48.