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A fisherman is interested in whether the distribution of fish caught in Green Valley Lake is the same as the distribution of fish caught in Echo Lake. Of the 191 randomly selected fish caught in Green Valley Lake, 105 were rainbow trout, 27 were other trout, 35 were bass, and 24 were catfish. Of the 293 randomly selected fish caught in Echo Lake, 115 were rainbow trout, 58 were other trout, 67 were bass, and 53 were catfish. Perform the hypothesis test at a 5% level of significance.

  • 3
  • Chi-Square with df = 3
  • 11.75
  • p-value = 0.0083

  • ii. Reject the null hypothesis.
    iv. There is sufficient evidence to conclude that the distribution of fish in Green Valley Lake is not the same as the distribution of fish in Echo Lake.

A plant manager is concerned her equipment may need recalibrating. It seems that the actual weight of the 15 oz. cereal boxes it fills has been fluctuating. The standard deviation should be at most 1 2 size 12{ { { size 8{1} } over { size 8{2} } } } {} oz. In order to determine if the machine needs to be recalibrated, 84 randomly selected boxes of cereal from the next day’s production were weighed. The standard deviation of the 84 boxes was 0.54. Does the machine need to be recalibrated?

  • 83
  • Chi-Square with df = 83
  • 96.81
  • p-value = 0.1426; There is a 0.1426 probability that the sample standard deviation is 0.54 or more.
  • Decision: Do not reject null; Conclusion: There is insufficient evidence to conclude that the standard deviation is more than 0.5 oz. It cannot be determined whether the equipment needs to be recalibrated or not.

Consumers may be interested in whether the cost of a particular calculator varies from store to store. Based on surveying 43 stores, which yielded a sample mean of $84 and a sample standard deviation of $12, test the claim that the standard deviation is greater than $15.

Isabella, an accomplished Bay to Breakers runner, claims that the standard deviation for her time to run the 7 ½ mile race is at most 3 minutes. To test her claim, Rupinder looks up 5 of her race times. They are 55 minutes, 61 minutes, 58 minutes, 63 minutes, and 57 minutes.

  • 4
  • Chi-Square with df = 4
  • 4.52
  • 0.3402
  • Decision: Do not reject null.

Airline companies are interested in the consistency of the number of babies on each flight, so that they have adequate safety equipment. They are also interested in the variation of the number of babies. Suppose that an airline executive believes the average number of babies on flights is 6 with a variance of 9 at most. The airline conducts a survey. The results of the 18 flights surveyed give a sample average of 6.4 with a sample standard deviation of 3.9. Conduct a hypothesis test of the airline executive’s belief.

The number of births per woman in China is 1.6 down from 5.91 in 1966 (Source World Bank, 6/5/12 ). This fertility rate has been attributed to the law passed in 1979 restricting births to one per woman. Suppose that a group of students studied whether or not the standard deviation of births per woman was greater than 0.75. They asked 50 women across China the number of births they had. Below are the results. Does the students’ survey indicate that the standard deviation is greater than 0.75?

# of births Frequency
0 5
1 30
2 10
3 5
  • 49
  • Chi-Square with df = 49
  • 54.37
  • p-value = 0.2774; If the null hypothesis is true, there is a 0.2774 probability that the sample standard deviation is 0.79 or more.
  • Decision: Do not reject null; Conclusion: There is insufficient evidence to conclude that the standard deviation is more than 0.75. It cannot be determined if the standard deviation is greater than 0.75 or not.

According to an avid aquariest, the average number of fish in a 20–gallon tank is 10, with a standard deviation of 2. His friend, also an aquariest, does not believe that the standard deviation is 2. She counts the number of fish in 15 other 20–gallon tanks. Based on the results that follow, do you think that the standard deviation is different from 2? Data: 11; 10; 9; 10; 10; 11; 11; 10; 12; 9; 7; 9; 11; 10; 11

The manager of "Frenchies" is concerned that patrons are not consistently receiving the same amount of French fries with each order. The chef claims that the standard deviation for a 10–ounce order of fries is at most 1.5 oz., but the manager thinks that it may be higher. He randomly weighs 49 orders of fries, which yields a mean of 11 oz. and a standard deviation of 2 oz.

  • σ 2 1 . 5 2 size 12{σ rSup { size 8{2} }<= left (1 "." 5 right ) rSup { size 8{2} } } {}
  • 48
  • Chi-Square with df = 48
  • 85.33
  • 0.0007
  • Decision: Reject null.

Try these true/false questions.

As the degrees of freedom increase, the graph of the chi-square distribution looks more and more symmetrical.

True

The standard deviation of the chi-square distribution is twice the mean.

False

The mean and the median of the chi-square distribution are the same if df = 24 size 12{ ital "df"="24"} {} .

False

In a Goodness-of-Fit test, the expected values are the values we would expect if the null hypothesis were true.

True

In general, if the observed values and expected values of a Goodness-of-Fit test are not close together, then the test statistic can get very large and on a graph will be way out in the right tail.

True

The degrees of freedom for a Test for Independence are equal to the sample size minus 1.

False

Use a Goodness-of-Fit test to determine if high school principals believe that students are absent equally during the week or not.

True

The Test for Independence uses tables of observed and expected data values.

True

The test to use when determining if the college or university a student chooses to attend is related to his/her socioeconomic status is a Test for Independence.

True

The test to use to determine if a six-sided die is fair is a Goodness-of-Fit test.

True

In a Test of Independence, the expected number is equal to the row total multiplied by the column total divided by the total surveyed.

True

In a Goodness-of Fit test, if the p-value is 0.0113, in general, do not reject the null hypothesis.

False

For a Chi-Square distribution with degrees of freedom of 17, the probability that a value is greater than 20 is 0.7258.

False

If df = 2 size 12{ ital "df"=2} {} , the chi-square distribution has a shape that reminds us of the exponential.

True

Questions & Answers

I'm interested in biological psychology and cognitive psychology
Tanya Reply
what does preconceived mean
sammie Reply
physiological Psychology
Nwosu Reply
How can I develope my cognitive domain
Amanyire Reply
why is communication effective
Dakolo Reply
Communication is effective because it allows individuals to share ideas, thoughts, and information with others.
effective communication can lead to improved outcomes in various settings, including personal relationships, business environments, and educational settings. By communicating effectively, individuals can negotiate effectively, solve problems collaboratively, and work towards common goals.
it starts up serve and return practice/assessments.it helps find voice talking therapy also assessments through relaxed conversation.
miss
Every time someone flushes a toilet in the apartment building, the person begins to jumb back automatically after hearing the flush, before the water temperature changes. Identify the types of learning, if it is classical conditioning identify the NS, UCS, CS and CR. If it is operant conditioning, identify the type of consequence positive reinforcement, negative reinforcement or punishment
Wekolamo Reply
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Wekolamo
because it helps many people around the world to understand how to interact with other people and understand them well, for example at work (job).
Manix Reply
Agreed 👍 There are many parts of our brains and behaviors, we really need to get to know. Blessings for everyone and happy Sunday!
ARC
A child is a member of community not society elucidate ?
JESSY Reply
Isn't practices worldwide, be it psychology, be it science. isn't much just a false belief of control over something the mind cannot truly comprehend?
Simon Reply
compare and contrast skinner's perspective on personality development on freud
namakula Reply
Skinner skipped the whole unconscious phenomenon and rather emphasized on classical conditioning
war
explain how nature and nurture affect the development and later the productivity of an individual.
Amesalu Reply
nature is an hereditary factor while nurture is an environmental factor which constitute an individual personality. so if an individual's parent has a deviant behavior and was also brought up in an deviant environment, observation of the behavior and the inborn trait we make the individual deviant.
Samuel
I am taking this course because I am hoping that I could somehow learn more about my chosen field of interest and due to the fact that being a PsyD really ignites my passion as an individual the more I hope to learn about developing and literally explore the complexity of my critical thinking skills
Zyryn Reply
good👍
Jonathan
and having a good philosophy of the world is like a sandwich and a peanut butter 👍
Jonathan
generally amnesi how long yrs memory loss
Kelu Reply
interpersonal relationships
Abdulfatai Reply
What would be the best educational aid(s) for gifted kids/savants?
Heidi Reply
treat them normal, if they want help then give them. that will make everyone happy
Saurabh
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Source:  OpenStax, Collaborative statistics (with edits: teegarden). OpenStax CNX. Jul 20, 2009 Download for free at http://legacy.cnx.org/content/col10561/1.3
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