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Hypothesis Testing of Two Means and Two Proportions: Review

Module by: Barbara Illowsky, Ph.D., Susan Dean. E-mail the authors

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Questions 1 – 3 refer to the following:

In a survey at Kirkwood Ski Resort the following information was recorded:

Table 1: Sport Participation by Age
  0 – 10 11 - 20 21 - 40 40+
Ski 10 12 30 8
Snowboard 6 17 12 5

Suppose that one person from of the above was randomly selected.

Exercise 1

Find the probability that the person was a skier or was age 11 – 20.

Solution

77 100 77 100 size 12{ { { size 8{"77"} } over { size 8{"100"} } } } {}

Exercise 2

Find the probability that the person was a snowboarder given he/she was age 21 – 40.

Solution

12 42 12 42 size 12{ { { size 8{"12"} } over { size 8{"42"} } } } {}

Exercise 3

Explain which of the following are true and which are false.

  • a. Sport and Age are independent events.
  • b. Ski and age 11 – 20 are mutually exclusive events.
  • c. P ( Ski and age 21 40 ) < P ( Ski age 21 40 ) P ( Ski and age 21 40 ) < P ( Ski age 21 40 ) size 12{P \( ital "Ski"+ ital "age""21" - "40" \) <P \( ital "Ski" \lline ital "age""21" - "40" \) } {}
  • d. P ( Snowboard orage 0 10 ) < P ( Snowboard age 0 10 ) P ( Snowboard orage 0 10 ) < P ( Snowboard age 0 10 ) size 12{P \( ital "Snowboardorage"0 - "10" \) <P \( ital "Snowboard" \lline ital "age"0 - "10" \) } {}

Solution

  • a. False
  • b. False
  • c. True
  • d. False

Exercise 4

The average length of time a person with a broken leg wears a cast is approximately 6 weeks. The standard deviation is about 3 weeks. Thirty people who had recently healed from broken legs were interviewed. State the distribution that most accurately reflects total time to heal for the thirty people.

Solution

N ( 180 , 16 . 43 ) N ( 180 , 16 . 43 ) size 12{N \( "180","16" "." "43" \) } {}

Exercise 5

The distribution for XX size 12{X} {} is Uniform. What can we say for certain about the distribution for X¯X¯ size 12{ {overline {X}} } {} when n=1n=1 size 12{n=1} {}?

  • A. The distribution for X¯X¯ size 12{ {overline {X}} } {} is still Uniform with the same mean and standard dev. as the distribution for XX size 12{X} {}.
  • B. The distribution for X¯X¯ size 12{ {overline {X}} } {}is Normal with the different mean and a different standard deviation as the distribution for XX size 12{X} {}.
  • C. The distribution for X¯X¯ size 12{ {overline {X}} } {} is Normal with the same mean but a larger standard deviation than the distribution for XX size 12{X} {}.
  • D. The distribution for X¯X¯ size 12{ {overline {X}} } {} is Normal with the same mean but a smaller standard deviation than the distribution for XX size 12{X} {}.

Solution

A

Exercise 6

The distribution for XX size 12{X} {} is uniform. What can we say for certain about the distribution for XX size 12{ Sum {X} } {} when n=50n=50 size 12{n=50} {}?

  • A. The distribution for XX size 12{ Sum {X} } {}is still uniform with the same mean and standard deviation as the distribution for XX size 12{X} {}.
  • B. The distribution for XX size 12{ Sum {X} } {} is Normal with the same mean but a larger standard deviation as the distribution for XX size 12{X} {}.
  • C. The distribution for XX size 12{ Sum {X} } {} is Normal with a larger mean and a larger standard deviation than the distribution for XX size 12{X} {}.
  • D. The distribution for XX size 12{ Sum {X} } {} is Normal with the same mean but a smaller standard deviation than the distribution for XX size 12{X} {}.

Solution

C

Questions 7 – 9 refer to the following:

A group of students measured the lengths of all the carrots in a five-pound bag of baby carrots. They calculated the average length of baby carrots to be 2.0 inches with a standard deviation of 0.25 inches. Suppose we randomly survey 16 five-pound bags of baby carrots.

Exercise 7

State the approximate distribution for X¯X¯ size 12{ {overline {X}} } {}, the distribution for the average lengths of baby carrots in 16 five-pound bags. X¯~X¯~ size 12{ {overline {X}} "~" } {}

Solution

N ( 2 . 25 16 ) N ( 2 . 25 16 ) size 12{N \( { { size 8{2 "." "25"} } over { size 8{ sqrt {"16"} } } } \) } {}

Exercise 8

Explain why we cannot find the probability that one individual randomly chosen carrot is greater than 2.25 inches.

Exercise 9

Find the probability that X¯X¯ size 12{ {overline {X}} } {} is between 2 and 2.25 inches.

Solution

0.5000

Questions 10 – 12 refer to the following:

At the beginning of the term, the amount of time a student waits in line at the campus store is normally distributed with a mean of 5 minutes and a standard deviation of 2 minutes.

Exercise 10

Find the 90th percentile of waiting time in minutes.

Solution

7.6

Exercise 11

Find the median waiting time for one student.

Solution

5

Exercise 12

Find the probability that the average waiting time for 40 students is at least 4.5 minutes.

Solution

0.9431

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