Question

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

Answer 1

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

Explanation:

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,
`H_0` : p
`\leq` 48% {means that the new blend is preferred by less than or equal to 48% of consumers}

Alternate Hypothesis,
`H_A` : 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. =
`\frac{\hat p-p}{\sqrt{(\hat p (1-\hat p))/(n) } }` ~ N(0,1)

where,
`\hat p` = sample proportion of tasters prefer new blend =
`(72)/(144)` = 0.50

n = sample of tasters = 144

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

= 0.48

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