Question

A newspaper reported that 15% of people say that some coffee shops are overpriced. The source of this information was a telephone survey of 100 adul a. Identify the population of interest in this study. b. Identify the sample for the study. c. Identify the parameter of interest in the study. d. Find and interpret a 95% confidence interval for the parameter of interest. a. Identify the population of interest. Choose the correct answer below. 100 adults adults coffee shops newspapers 15% of adults a telephone survey b. Identify the sample. Choose the correct answer below. 15% of adults newspapers a telephone survey 100 adults coffee shops adults c. Identify the parameter of interest. Choose the correct answer below. A. p, the population proportion of adults who say that some coffee shops are overpriced O B. p, the sample proportion of adults who say that some coffee shops are overpriced O C. x, the sample mean number of adults who say that some coffee shops are overpriced O D. 62, the population variance of adults who say that some coffee shops are overpriced d. The 95% confidence interval for the parameter of interest is ( ). (Round to two decimal places as needed.)

Answer 1

a. Here, the population of interests are adults.

b. Since the information is based on the telephone survey of 100 adults, thus the sample is 100 adults.

c. The parameter of interest is the population proportion of adults who say that some coffee shops are overpriced. Thus option (A) is correct.

d. (1-alpha)100% confidence interval for population proportion is given by the formula

p' +- z(alpha/2) * sqrt(p' (1-p')/n)

where p' is the sample proportion given as 15% or 0.15

For 95% confidence interval we have z(alpha/2) = 1.96

Again p'(1-p') /n = 0.15*0.85 / 100

= .1275/100 = .001275

Thus sqrt(p' (1-p')/n) = sqrt ( 0.001275) = 0.0357071421

Again z(alpha/2)* sqrt(p'(1-p')/n) = 1.96*0.0357071421

= 0.0699859985

Thus the lower confidence limit is given by = p' - 0.0699859985= 0.15- 0.0699859985= 0.08 ( two decimal places)

and the upper confidence limit is given by = p'+ 0.0699859985 = 0.15+ 0.0699859985= 0.22 ( two decimal places)

Thus the 95% confidence interval for the population proportion is given as (0.08, 0.22).