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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

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Source:  OpenStax, Collaborative statistics homework book: custom version modified by r. bloom. OpenStax CNX. Dec 23, 2009 Download for free at http://legacy.cnx.org/content/col10619/1.2
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