Question

A newspaper reported that 55% of people say that some coffee shops are overpriced. The source of this information was a telephone survey of 100 adults a. ldentify the population of interest in this study. b. Identify the sample for the study c. Identify the parameter of interest in the study Find and interpret a 90% confidence interval for the parameter of interest. a. Identify the population of interest. Choose the correct answer below. d newspapers ( 55% of adults O adults b. Identify the sample. Choose the correct answer below. O O O coffee shops a telephone survey 100 adults O coffee shops O 100 adults O adults O newspapers O a telephone survey 55% of adults c. Identify the parameter of interest. Choose the correct answer below. O A. p, the population proportion of adults who say that some coffee shops are overpriced B. x, the sample mean number of adults who say that some coffee shops are overpriced C. ơ2, the population variance of adults who say that some coffee shops are overpriced p, the sample proportion 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. 0 A. We are confident that 90% of the population is outside the interval for the parameter of interest. 0 B. There is a 90% chance that the value of the parameter of interest is outside the interval c. we are confident that 90% of the population is described by the interval for the parameter of interest. 0 D. We are 90% confident that the parameter of interest lies in the confidence interval

Answer 1

a) Here we select 100 adults from the population of all adults.

So correct choice of this part is " Adults".

b) Here we take 100 adults from all adults. So correct option is " 100 adults".

c) The correct option is A)

d) Formula of confidence interval for population proportion is

Where E is called margin of error

Formula of E is

It is given that ; c = confidence level = 0.90

so that level of significance = = 1 - c = 1 - 0.90 = 0.10

this implies that /2 = 0.10/2 = 0.05

So we want to find Z_{/2} such that  

P(Z >Z_{/2}) = 0.05

Therefore ,

P(Z < Z_{/2}) = 1 - 0.05 = 0.95

The general excel command to find critical z value is

"=NORMSDIST(probability)"

Here probability = 0.95

So that critical Z_{/2} is = "=NORMSINV(0.95)" = 1.645

also n = 100 and = 0.55

Put this value in equation (1) so we get

So E = 0.081838

Lower limit = - E = 0.55 - 0.081838 = 0.468162 = 0.47

Upper Limit = + E = 0.44 + 0.081838 = 0.631838 = 0.63

So that 90% confidence interval for population proportion P is (0.47 , 0.63)

The correct choice is