A newspaper reported that 20% of people say that some coffee shops are overprice

Question

A newspaper reported that 20% of people say that some coffee shops are overpriced. The source of this information was a telephone survey of 100 adults. 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 90% confidence interval for the parameter of interest. a. Identify the population of interest. Choose the correct answer below. 20% of adults O 100 adults O adults O O coffee shops a telephone survey b. ldentify the sample. Choose the correct answer below. O coffee shops O 20% of adults O a telephone survey c. Identify the parameter of interest. Choose the correct answer below. O newspapers O 100 adults p, the sample proportion of adults who say that some coffee shops are overpriced O B. p. the population proportion of adults who say that some coffee shops are overpriced ° C. 2, the population variance of adults who say that some coffee shops are overpriced O D. x, the sample mean number of adults who say that some coffee shops are overpriced d. The 90% confidence interval for the parameter of interest is ( (Round to two decimal places as needed.) Interpret this confidence interval. Choose the correct answer below. OD. A. 0 B. C. 0 D. We are confident that 90% of the population is described by the interval for the parameter of interest. We are confident that 90% of the population is outside the interval for the parameter of interest. We are 90% confident that the parameter of interest lies in the confidence interval. There is a 90% chance that the value of the parameter of interest is outside the interval.

Explanation / Answer

Hello,

a) 20% of adults

here 20% are saying the coffee shops are overprized.

b) 100 people

survey is conducted on the 100 peoples.

c) b is correct

Because we are always talking about the population, proving that thing we take a sample.

d)

Standard error of the mean = SEM = x(N-x)/N3 = 0.040

= (1-CL)/2 = 0.050

Standard normal deviate for = Z = 1.645

Proportion of positive results = P = x/N = 0.200

Lower bound = P - (Z*SEM) = 0.134

Upper bound = P + (Z*SEM) = 0.266

c answer is correct

Thanks,

Thanks

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