Question

In: Statistics and Probability

One of the authors came across an article (USA Today, 2008) that said that on average...

One of the authors came across an article (USA Today, 2008) that said that on average Americans have visited 16 states in the United States. In a survey of 50 students in her introductory statistics class, she found the average number of states the students had visited to be 9.48 and the standard deviation to be 7.13. The data were not strongly skewed.​

1. Identify the observational unit for this study.

a. students

b. Americans

c. number of states

2. Identify the variable of interest and whether it is categorical or quantitative.

a. number of U.S. states visited, categorial

b. number of students, quantitative

c. number of students, categorial

d. number of U.S. states visited, quantitative

3. Regardless of your answer to part (c), state the null and the alternative hypotheses in symbols, to test whether the average number of states all students at the author’s school have visited is different from 16.

a. H0: mu = 16, Ha: mu ≠ 16

b. H0: mu = 16, Ha: mu > 16

c. H0: mu = 16, Ha: mu < 16

4. Using the 2SD approach to find a 95% confidence interval for the average number of states all students at the author’s school have visited. Round to two decimal places

Solutions

Expert Solution

1. The researcher is interested in finding out the average no. of states in US that are visited by Americans.

The unit that is being observed to arrive at the required conclusion is an American. The observational unit for this study is Ans: b. Americans

2. The variable of interest is number of U.S. states visited. Since the observations recorded for this variable are numeric in nature, the variable is quantitative in nature.

Ans: d. number of U.S. states visited, quantitative

3. To test whether the average number of states all students at the author’s school have visited,

The null hypothesis is always a negation statement/null of what we wish to test. Here, we wish to test that the average is different from 16.Hence, the null would be that the average is 16.

Ans: a. H0: mu = 16, Ha: mu ≠ 16

4. A  95% confidence interval for the average number of states all students at the author’s school have visited can be computed using the formula:

= (7.46,11.50)

The 95% confidence interval for the average number of states all students at the author’s school have visited is (7.46,11.50)


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