Question

3. In 2000, a sample of 209 people aged 18-30 found that they spent an average of 6.75 hrs/week on the internet. In 2006, a sample of 541 people aged 18-30 spent an average of 7.34 hrs/week on the internet. Has the average increased?

Use correct notation where appropriate.
For each of the following situations,
a) What is/are the population(s) in the study?
b) Does the scenario involve 1 or 2 parameters to be estimated/tested/compared?
c) What is/are the parameter(s) to be estimated/tested/compared? Be specific and use the appropriate notation.
d) What is/are the variable(s) involved. Is each numerical or categorical?
e) Is it most appropriate to create a confidence interval or conduct a hypothesis test? If a hypothesis test should be conducted, state the
hypotheses.
f) Will the critical value or test statistic be a z-value or t-value?

Answer 1

(a) The populations in study are the people aged 18 - 30 who spent on the internet in the years 2000 and 2006.

(b) The scenario involves 2 parameters to be compared.

(c) The parameters to be compared are p1, which is the proportion of people aged 18 - 30 who spent on the internet in 2000 and p2, the proportion of people 18-30 who spent on the internet in 2006. The comparison to be made is if p1 is > p2.

(d) The variables involved are

(i) Amount Spent - Numerical

(ii) The year in which it was spent - Categorical

(iii) The age group - Categorical.

(e) It is better to do a Hypothesis test as we are doing a right tailed test (CI is 2 tailed)

The Hypothesis is:

H0: p1 = p2: The proportion of people aged 18 - 30 who shopped on the internet in 2002 is equal to the proportion in 2006.

Ha: p1 > p2: The proportion of people aged 18 - 30 who shopped on the internet in 2002 is greater then the proportion of those in 2006.

(f) The critical value will be a z value, as the following conditions are met

(i) The sampling is random

(ii) The samples are independent of each other

(iii) n *
P1
, n * (1 -
P1
), n * $\hat{p_2}$ and n * (1 - $\hat{p_2}$) are all greater than 10.