Question

In a recent Dunkin Donuts (DD) commercial, it states that 58% of Americans prefer to drink their coffee. Of the five hundred randomly selected people who were asked if they preferred DD coffees, 325 said they did. As a DD fanatic, you are interested in determining if the proportion of Americans who prefer to drink DD coffee is actually more than what was advertised. (alpha = 0.05) Show all your work. State the null and alternative hypotheses. Define the parameter.

Answer 1

We conclude that the proportion of Americans who prefer to drink DD coffee is actually more than what was advertised.

Explanation:

We are given that in a recent Dunkin Donuts (DD) commercial, it states that 58% of Americans prefer to drink their coffee.

Of the five hundred randomly selected people who were asked if they preferred DD coffees, 325 said they did.

Let p = proportion of Americans who prefer to drink DD coffee

SO, Null Hypothesis,
`H_0` : p
`\leq` 58% {means that the proportion of Americans who prefer to drink DD coffee is actually less than or equal to what was advertised}

Alternate Hypothesis,
`H_A` : p > 58% {means that the proportion of Americans who prefer to drink DD coffee is actually more than what was advertised}

The test statistics that will be used here is 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 people who prefer to drink DD coffee in a sample of 500 people =
`(325)/(500)` = 65% or 0.65

n = sample of people = 500

So, test statistics =
`\frac{0.65-0.58}{{\sqrt{(0.65(1-0.65))/(500) } } } }`

= 3.282

Now at 0.05 significance level, the z table gives critical value of 1.6449 for right-tailed test. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the proportion of Americans who prefer to drink DD coffee is actually more than what was advertised.