Preparing to Estimate a Population Proportion Assignment
Identify the z* critical value for constructing a confidence interval for one proportion using these confidence levels. You can reference the t distribution table to answer this question. Enter each answer to 3 decimal places.
-1.645 -1.960 -2.576
What is the z* critical value for constructing a 94% confidence interval for one proportion? This value is not listed on the t distribution table, so you will need to use this z-table to determine the value. Round to 2 decimal places.
-1.88
A doctor would like to estimate the proportion of adults who eat an apple a day. To learn more about this, she selects a random sample of 200 adults and finds that 24 of them eat an apple a day. She would like to construct a 98% confidence interval for the true proportion of adults who eat an apple a day.
-met -met -met -yes
A random sample of 100 magazine subscribers finds that 38 do not read the magazine they subscribe to. We would like to construct a 99% confidence interval for the true proportion of all magazine subscribers who do not read the magazine they subscribe to.
-met -met -met -yes
Companies love selling gift cards because 1.5% of gift cards have historically gone unused. Gift cards are tracked using a bar code, so their usage is easily recorded. A random sample of 500 gift cards that were purchased more than 1 year ago are randomly selected and it is found that 20 of them are unused. We would like to construct a 99% confidence interval for the true proportion of gift cards sold1 year ago that are still currently unused.
-met -met -met -yes
A credit card company would like to estimate the proportion of their customers who have at least $10,000 in credit card debt. They select a random sample of 50 of their customers and find that 42 of them have at least $10,000 in credit card debt. They would like to construct a 95% confidence interval for the true proportion of their customers who have at least $10,000 in credit card debt.
-met -met -not met -no
A college professor randomly selects 10 of the 28 students in his statistics class and finds that 8 of them have part-time jobs in addition to being a full-time student. He would like to construct a 90% confidence interval for the true proportion of students in his class who have part-time jobs in addition to being a full-time student.
-met -not met -not met -no
A random sample of 240 doctors revealed that 108 are satisfied with the current state of US health care. The conditions for inference are met. A 90% confidence interval for the true proportion of all US doctors who are satisfied with the current state of US health care is (0.397, 0.503).
90% of all doctors are satisfied with the current state of US health care. Between 39.7% and 50.3% of the sample of doctors are satisfied with the current state of US health care. We are 90% confident that the interval from 0.397 to 0.503 captures p = the true proportion of all US doctors who are satisfied with the current state of US health care.(correct) A majority of the doctors in the sample are satisfied with the current state of US health care.
A random sample of 240 doctors revealed that 108 are satisfied with the current state of US health care. The conditions for inference are met. Using z* = 1.645, which expression gives a 90% confidence interval for the true proportion of all US doctors who are satisfied with the current state of US health care?
c
