Question

47. The American Society for Microbiology (ASM) and the Soap and Detergent Association (SDA) jointly commissioned two seperate studies, both of which were conducted by Harris Interactive. IN one of the studies, 1001 adults were interviewed by the telephone and asked about their handwashing habits. In the telephone inverviews, 921 of the adults said they always wash their hands in public restrooms. In the other study, the hand-washing behavior of 6076 adults was inconspicuously observed within public restrooms in four U.S. cities and 4670 of the 6076 adults were observed washing their hands.

a. In the telephone survey, what is the variable of interest? Is it qualitative or quantitative?

b. What is the sample in the telephone survery? What is the population to which this study applies?

c. Verify that the requirements for constructing a confidence interval for the population proportion of adults who say they always wash their hands in public restrooms are satisified.

d. Using the results from the telephone interviews, construct a 95% confidence interval for the proportion of adults who say they always was their hands in public restrooms.

e. In the study where hand-washing behavior was observed, what is the variable of interest? Is it qualitative or quantitative?

f. We are told that 6076 adults were inconspicuously ob served, but wer are not told how these adults were selected. We know randomness is a key ingredient in statistical studies that allows us to generalize results from a sample to a population. Suggest some ways randomness might have been used to select individuals in this study.

g. Verify the requirements for constructing a confidence interval for the population proportion of adults who actually washed their hands while in a public restroom.

h. Using the results from the observational study, construct a 95% confidence interval for the proportion of adults who wash their hands in public restrooms.

i. Based on your findings in parts (a) through (h), what might you conclude about the proportion of adults who say they always wash their hands versus the proportion of adults who actually wash their hands in public restrooms?

j. Cite some sources of variability in both studies.

Answer 1

a. In the telephone survey, what is the variable of interest? Is it qualitative or quantitative?

In the telephone survey, the variable of interest is given as the hand washing habit. This variable is qualitative in nature.

b. What is the sample in the telephone survery? What is the population to which this study applies?

The sample in the telephone survey is given as the 1001 adults who were interviewed by the telephone and asked about their hand washing habits. The population to which this study applies is given as all adults in the US.

c. Verify that the requirements for constructing a confidence interval for the population proportion of adults who say they always wash their hands in public restrooms are satisfied.

The main assumption of random sample is satisfied because about 1001 adults were selected randomly and then interviewed by the telephone. Also, the variable under study is binary in nature, that is, it results only in responses as ‘yes’ or ‘no’. So, we can use confidence interval for the population proportion.

d. Using the results from the telephone interviews, construct a 95% confidence interval for the proportion of adults who say they always was their hands in public restrooms.

Here, we have to find the 95% confidence interval for the proportion of adults who say they always wash their hands in public restrooms.

Formula for confidence interval is given as below:

Confidence interval for Population Proportion

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is sample proportion, n is sample size, and Z is critical value which we have to find by using z-table.

From given information, we have

Sample size = n = 1001

Number of successes = X = 921

Confidence level = 95%

Critical Z value = 1.96 (by using z-table)

Sample proportion = X/n = 921/1001 = 0.92007992

Confidence interval = 0.92007992 ± 1.96*sqrt(0.92007992*(1 - 0.92007992)/1001)

Confidence interval = 0.92007992 ± 1.96* 0.0086

Confidence interval = 0.92007992 ± 0.0168

Lower limit = 0.92007992 - 0.0168 =0.90328

Upper limit = 0.92007992 + 0.0168 = 0.93688

Confidence interval = (0.90328, 0.93688)