Question

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?

Answer 1

Answer:

There is enough evidence to conclude that mean annual consumption of chocolate is higher in Hershey than it is throughout the United States.

Step-by-step explanation:

We are given 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

First, we design the null and the alternate hypothesis

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

We use one-tailed t test to perform this hypothesis.

Formula:

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

Putting all the values, we have

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

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

Since,                          

The calculated t statistic is greater that critical t value, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis

Thus, there is enough evidence to conclude that mean annual consumption of chocolate is higher in Hershey than it is throughout the United States.

Answer 2

Using the t-distribution, it is found that since the p-value of the test is of 0.0173 < 0.05, the sample results support the conclusion that mean annual consumption of chocolate is higher in Hershey than it is throughout the United States.

At the null hypothesis, it is tested if the mean in Hershey is the same as in the rest of the US, that is, of 9.5 pounds, hence:

$H_0: \mu = 9.5$

At the alternative hypothesis, it is tested if the mean is greater, that is:

$H_1: \mu > 9.5$

We have the standard deviation for the sample, thus, the t-distribution is used. The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

• $\overline{x}$ is the sample mean.
• $\mu$ is the value tested at the null hypothesis.
• s is the standard deviation of the sample.
• n is the sample size.

For this problem, the values of the parameters are: $\overline{x} = 10.05, \mu = 9.5, s = 1.5, n = 36$

The value of the test statistic is:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{10.05 - 9.5}{\frac{1.5}{\sqrt{36}}}$

$t = 2.2$

The p-value of the test is found using a right-tailed test, as we are testing if the mean is greater than a value, with t = 2.2 and 36 - 1 = 35 df.

Using a t-distribution calculator, this p-value is of 0.0173.

Since the p-value of the test is of 0.0173 < 0.05, the sample results support the conclusion that mean annual consumption of chocolate is higher in Hershey than it is throughout the United States.

A similar problem is given at https://brainly.com/question/25461582