The United States ranks ninth in the world in per capita chocolate consumption; Forbes reports that the average American eats 9.

Question

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The United States ranks ninth in the world in per capita chocolate consumption; Forbes reports that the average American eats 9.

5 pounds of chocolate annually. Suppose you are curious whether chocolate consumption is higher in Hershey, Pennsylvania, the location of the Hershey Company’s corporate headquarters. A sample of 36 individuals from the Hershey area showed a sample mean annual consumption of 10.05 pounds and a standard deviation of s= 1.5 pounds. Using a=.05, do the sample results support the conclusion that mean annual consumption of chocolate is higher in Hershey than it is throughout the United States?

Mathematics

1 answer:

Leona [9.2K] · Answer 1 · 2026-03-25T12:38:32+03:00

Answer:

Sufficient evidence exists to determine that the average yearly chocolate consumption in Hershey surpasses that of the entire United States.

Step-by-step explanation:

The information provided includes the following in the question:

Population mean, μ = 9.5 pounds

Sample mean, $\bar{x}$ = 10.05 pounds

Sample size, n = 36

Alpha, α = 0.05

Sample standard deviation, s = 1.5 pounds

To begin with, we formulate the null and alternative hypotheses

$H_{0}: \mu = 9.5\text{ pounds}\\H_A: \mu > 9.5\text{ pounds}$

We apply a one-tailed t test to evaluate this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

By substituting all the values, we obtain

$t_{stat} = \displaystyle\frac{10.05 - 9.5}{\frac{1.5}{\sqrt{36}} } = 2.2$

At this point, $t_{critical} \text{ at 0.05 level of significance, 35 degree of freedom } = 1.6895$

Given that,

The calculated t statistic exceeds the critical t value, we reject the null hypothesis instead of accepting it. We adopt the alternative hypothesis

Consequently, there is sufficient evidence to assert that the average annual chocolate consumption in Hershey is greater than that in the United States as a whole.