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11.15: Chapter 11 Solution (Practice + Homework)

  • Page ID
    6140
  • 1.

    mean = 25 and standard deviation = 7.0711

    3.

    when the number of degrees of freedom is greater than 90

    5.

    \(df = 2\)

    6.

    a test of a single variance

    8.

    a left-tailed test

    10.

    \(H_0: \sigma^2 = 0.812\);

    \(H_a: \sigma^2 > 0.812\).

    12.

    a test of a single variance

    16.

    a goodness-of-fit test

    18.

    3

    20.

    2.04

    21.

    We decline to reject the null hypothesis. There is not enough evidence to suggest that the observed test scores are significantly different from the expected test scores.

    23.

    \(H_0\): the distribution of AIDS cases follows the ethnicities of the general population of Santa Clara County.

    25.

    right-tailed

    27.

    2016.136

    28.

    • Graph: Check student’s solution.
    • Decision: Cannot accept the null hypothesis.
    • Reason for the Decision: Calculated value of test statistics is either in or out of the tail of the distribution.
    • Conclusion (write out in complete sentences): The make-up of AIDS cases does not fit the ethnicities of the general population of Santa Clara County.

    30.

    a test of independence

    32.

    a test of independence

    34.

    8

    36.

    6.6

    39.

    Smoking level per dayAfrican AmericanNative HawaiianLatinoJapanese AmericansWhiteTotals
    1-109,8862,74512,8318,3787,65041,490
    11-206,5143,0624,93210,6809,87735,065
    21-301,6711,4191,4064,7156,06215,273
    31+7597888002,3053,9708,622
    Totals18,8308,01419,96926,07827,55910,0450

    Table 11.54

    41.

    Smoking level per dayAfrican AmericanNative HawaiianLatinoJapanese AmericansWhite
    1-107777.573310.118248.0210771.2911383.01
    11-206573.162797.526970.769103.299620.27
    21-302863.021218.493036.203965.054190.23
    31+1616.25687.871714.012238.372365.49

    Table 11.55

    43.

    10,301.8

    44.

    right

    46.

    1. Cannot accept the null hypothesis.
    2. Calculated value of test statistics is either in or out of the tail of the distribution.
    3. There is sufficient evidence to conclude that smoking level is dependent on ethnic group.

    48.

    test for homogeneity

    50.

    test for homogeneity

    52.

    All values in the table must be greater than or equal to five.

    54.

    3

    57.

    a goodness-of-fit test

    59.

    a test for independence

    61.

    Answers will vary. Sample answer: Tests of independence and tests for homogeneity both calculate the test statistic the same way \(\sum_{(i j)} \frac{(O-E)^{2}}{E}\). In addition, all values must be greater than or equal to five.

    63.

    true

    65.

    false

    67.

    225

    69.

    \(H_0: \sigma^2 \leq 150\)

    71.

    36

    72.

    Check student’s solution.

    74.

    The claim is that the variance is no more than 150 minutes.

    76.

    a Student's \(t\)- or normal distribution

    78.

    1. \(H_0: \sigma = 15\)
    2. \(H_a: \sigma > 15\)
    3. \(df = 42\)
    4. chi-square with \(df = 42\)
    5. test statistic = 26.88
    6. Check student’s solution.
      • Alpha = 0.05
      • Decision: Cannot reject null hypothesis.
      • Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
      • Conclusion: There is insufficient evidence to conclude that the standard deviation is greater than 15.

    80.

    1. \(H_0: \sigma \leq 3\)
    2. Ha: \sigma > 3\)
    3. \(df = 17\)
    4. chi-square distribution with \(df = 17\)
    5. test statistic = 28.73
    6. Check student’s solution.
      • Alpha: 0.05
      • Decision: Cannot accept the null hypothesis.
      • Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
      • Conclusion: There is sufficient evidence to conclude that the standard deviation is greater than three.

    82.

    1. \(H_0: \sigma = 2\)
    2. Ha: \sigma \neq 2\)
    3. \(df = 14\)
    4. chi-square distiribution with \(df = 14\)
    5. chi-square test statistic = 5.2094
    6. Check student’s solution.
      • Alpha = 0.05
      • Decision: Cannot accept the null hypothesis
      • Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
      • Conclusion: There is sufficient evidence to conclude that the standard deviation is different than 2.

    84.

    1. \(H_0 : \sigma^2 = 252\)
    2. \(Ha : \sigma^2 > 252\)
    3. \(df = n – 1 = 7.\)
    4. test statistic: \(x^{2}=x_{7}^{2}=\frac{(n-1) s^{2}}{25^{2}}=\frac{(8-1)(34.29)^{2}}{25^{2}}=13.169\);
      • Alpha = 0.05
      • Decision: Cannot accept the null hypothesis
      • Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
      • Conclusion: At the 5% level, there is insufficient evidence to conclude that the variance is more than 625.

    87.

    Marital statusPercentExpected frequency
    Never married31.3125.2
    Married56.1224.4
    Widowed2.510
    Divorced/Separated10.140.4

    Table 11.56

    1. The data fits the distribution.
    2. The data does not fit the distribution.
    3. 3
    4. chi-square distribution with \(df = 3\)
    5. 19.27
    6. 0.0002
    7. Check student’s solution.
      • Alpha = 0.05
      • Decision: Cannot accept null hypothesis at the 5% level of significance
      • Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
      • Conclusion: Data does not fit the distribution.

    89.

    1. \(H_0\): The local results follow the distribution of the U.S. AP examinee population
    2. \(H_a\): The local results do not follow the distribution of the U.S. AP examinee population
    3. \(df = 5\)
    4. chi-square distribution with \(df = 5\)
    5. chi-square test statistic = 13.4
    6. Check student’s solution.
      • Alpha = 0.05
      • Decision: Cannot accept null when \(\alpha = 0.05\)
      • Reason for Decision: Calculated value of test statistics is either in or out of the tail of the distribution.
      • Conclusion: Local data do not fit the AP Examinee Distribution.
      • Decision: Do not reject null when \(\alpha = 0.01\)
      • Conclusion: There is insufficient evidence to conclude that local data do not follow the distribution of the U.S. AP examinee distribution.

    91.

    1. \(H_0\): The actual college majors of graduating females fit the distribution of their expected majors
    2. \(H_a\): The actual college majors of graduating females do not fit the distribution of their expected majors
    3. \(df = 10\)
    4. chi-square distribution with \(df = 10\)
    5. test statistic = 11.48
    6. Check student’s solution.
      • Alpha = 0.05.
      • Decision: Cannot reject null when \(\alpha = 0.05\) and \(\alpha = 0.01\).
      • Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
      • Conclusion: There is insufficient evidence to conclude that the distribution of actual college majors of graduating females fits the distribution of their expected majors.

    94.

    true

    96.

    false

    98.

    1. \(H_0\): Surveyed obese fit the distribution of expected obese
    2. \(H_a\): Surveyed obese do not fit the distribution of expected obese
    3. \(df = 4\)
    4. chi-square distribution with \(df = 4\)
    5. test statistic = 54.01
    6. Check student’s solution.
      • Alpha: 0.05
      • Decision: Cannot accept the null hypothesis.
      • Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
      • Conclusion: At the 5% level of significance, from the data, there is sufficient evidence to conclude that the surveyed obese do not fit the distribution of expected obese.

    100.

    1. \(H_0\): Car size is independent of family size.
    2. \(H_a\): Car size is dependent on family size.
    3. \(df = 9\)
    4. chi-square distribution with \(df = 9\)
    5. test statistic = 15.8284
    6. Check student’s solution.
      • Alpha: 0.05
      • Decision: Cannot reject the null hypothesis.
      • Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
      • Conclusion: At the 5% significance level, there is insufficient evidence to conclude that car size and family size are dependent.

    102.

    1. \(H_0\): Honeymoon locations are independent of bride’s age.
    2. \(H_a\): Honeymoon locations are dependent on bride’s age.
    3. \(df = 9\)
    4. chi-square distribution with \(df = 9\)
    5. test statistic = 15.7027
    6. Check student’s solution.
      • Alpha: 0.05
      • Decision: Cannot reject the null hypothesis.
      • Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
      • Conclusion: At the 5% significance level, there is insufficient evidence to conclude that honeymoon location and bride age are dependent.

    104.

    1. \(H_0\): The types of fries sold are independent of the location.
    2. \(H_a\): The types of fries sold are dependent on the location.
    3. \(df = 6\)
    4. chi-square distribution with \(df = 6\)
    5. test statistic =18.8369
    6. Check student’s solution.
      • Alpha: 0.05
      • Decision: Cannot accept the null hypothesis.
      • Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
      • Conclusion: At the 5% significance level, There is sufficient evidence that types of fries and location are dependent.

    106.

    1. \(H_0\): Salary is independent of level of education.
    2. \(H_a\): Salary is dependent on level of education.
    3. \(df = 12\)
    4. chi-square distribution with \(df = 12\)
    5. test statistic = 255.7704
    6. Check student’s solution.
      • Alpha: 0.05
      • Decision: Cannot accept the null hypothesis.
      • Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
      • Conclusion: At the 5% significance level, there is sufficient evidence to conclude that salary and level of education are dependent.

    108.

    true

    110.

    true

    112.

    1. \(H_0\): Age is independent of the youngest online entrepreneurs’ net worth.
    2. \(H_a\): Age is dependent on the net worth of the youngest online entrepreneurs.
    3. \(df = 2\)
    4. chi-square distribution with \(df = 2\)
    5. test statistic = 1.76
    6. Check student’s solution.
      • Alpha: 0.05
      • Decision: Cannot reject the null hypothesis.
      • Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
      • Conclusion: At the 5% significance level, there is insufficient evidence to conclude that age and net worth for the youngest online entrepreneurs are dependent.

    114.

    1. \(H_0\): The distribution for personality types is the same for both majors
    2. \(H_a\): The distribution for personality types is not the same for both majors
    3. \(df = 4\)
    4. chi-square with \(df = 4\)
    5. test statistic = 3.01
    6. Check student’s solution.
      • Alpha: 0.05
      • Decision: Cannot reject the null hypothesis.
      • Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
      • Conclusion: There is insufficient evidence to conclude that the distribution of personality types is different for business and social science majors.

    116.

    1. \(H_0\): The distribution for fish caught is the same in Green Valley Lake and in Echo Lake.
    2. \(H_a\): The distribution for fish caught is not the same in Green Valley Lake and in Echo Lake.
    3. 3
    4. chi-square with \(df = 3\)
    5. 11.75
    6. Check student’s solution.
      • Alpha: 0.05
      • Decision: Cannot accept the null hypothesis.
      • Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
      • Conclusion: There is evidence to conclude that the distribution of fish caught is different in Green Valley Lake and in Echo Lake

    118.

    1. \(H_0\): The distribution of average energy use in the USA is the same as in Europe between 2005 and 2010.
    2. \(H_a\): The distribution of average energy use in the USA is not the same as in Europe between 2005 and 2010.
    3. \(df = 4\)
    4. chi-square with \(df = 4\)
    5. test statistic = 2.7434
    6. Check student’s solution.
      • Alpha: 0.05
      • Decision: Cannot reject the null hypothesis.
      • Reason for decision: Calculated value of test statistics is either in or out of the tail of the distribution.
      • Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the average energy use values in the US and EU are not derived from different distributions for the period from 2005 to 2010.

    120.

    1. \(H_0\): The distribution for technology use is the same for community college students and university students.
    2. \(H_a\): The distribution for technology use is not the same for community college students and university students.
    3. 2
    4. chi-square with \(df = 2\)
    5. 7.05
    6. p-value = 0.0294
    7. Check student’s solution.
      • Alpha: 0.05
      • Decision: Cannot accept the null hypothesis.
      • Reason for decision: p-value < alpha
      • Conclusion: There is sufficient evidence to conclude that the distribution of technology use for statistics homework is not the same for statistics students at community colleges and at universities.

    122.

    1. The test statistic is always positive and if the expected and observed values are not close together, the test statistic is large and the null hypothesis will be rejected.
    2. Testing to see if the data fits the distribution “too well” or is too perfect.