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Practice calculating and interpreting the different types of confidence interval

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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 13th
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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