Practice calculating and interpreting the different types of confidence interval
22. You Explain It! Superstition: A USA Today/Gallup poll asked 1,006 adult Americans how much it would bother them to stay in a room on the 13th ﬂoor of a hotel. Interestingly, 13% said it would bother them. The margin of error was 3 percentage points with 95% conﬁdence. Which of the following represents a reasonable interpretation of the survey results? For those not reasonable, explain the ﬂaw. |
(a) We are 95% conﬁdent that the proportion of adult Americans who would be bothered to stay in a room on the 13th ﬂoor is between 0.10 and 0.16. |
Answer: |
(b) We are between 92% and 98% conﬁdent that 13% of adult Americans would be bothered to stay in a room on the 13^{th} ﬂoor. |
Answer: |
(c) In 95% of samples of adult Americans, the proportion who would be bothered to stay in a room on the 13th ﬂoor is between 0.10 and 0.16. |
Answer: |
(d) We are 95% conﬁdent that 13% of adult Americans would be bothered to stay in a room on the 13th ﬂoor. |
Answer: |
Section 9.2 |
8. (a) Find the t-value such that the area in the right tail is 0.02 with 19 degrees of freedom. |
Answer: |
(b) Find the t-value such that the area in the right tail is 0.10 with 32 degrees of freedom. |
Answer: |
(c) Find the t-value such that the area left of the t-value is 0.05 with 6 degrees of freedom. [Hint: Use symmetry.] |
Answer: |
(d) Find the critical t-value that corresponds to 95% conﬁdence. Assume 16 degrees of freedom. |
Answer: |
Section 9.3 |
7. Aggravated Assault: In a random sample of 40 felons convicted of aggravated assault, it was determined that the mean length of sentencing was 54 months, with a standard deviation of 8 months. Construct and interpret a 95% conﬁdence interval for the mean length of sentence for an aggravated assault conviction. Source: Based on data from the U.S. Department of Justice. |
Answer: |
12. Theme Park Spending: In a random sample of 40 visitors to a certain theme park, it was determined that the mean amount of money spent per person at the park (including ticket price) was $93.43 per day with a standard deviation of $15. Construct and interpret a 99% conﬁdence interval for the mean amount spent daily per person at the theme park. |
Answer: |
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