Note:
  • Make sure to describe your evidence in Step 3 for why you are performing a chi-square test. It is not adequate to simply say “because there is a categorical response variable and two or more groups are being tests.” At least say what the response variable is and what the groups are.
  • When testing the assumptions (Step 5), make sure to explicitly say that there are more than five in each cell of the expected table. It is not adequate to simply make the expected table.
  • In hand-calculation questions (e.g., hurricanes) make sure to put lower.tail=FALSE in distrib() when calculating the p-value (Step 8) in a chi-square problem. This is true for EVERY chi-square problem.
  • You values may differ slightly from mine as I calculate everything in R that holds many decimal places on intermediate values.

Hurricane Strengths

  1. α=0.10.
  2. H0: “The distribution of hurricanes into the strength categories is the same for both time periods” versus HA: “The distribution of hurricanes into the strength categories is NOT the same for both time periods”
  3. A chi-square test is required because (i) a categorical response variable with three levels (strength category) from (ii) two groups or populations (time periods) was recorded.
  4. This is an observational study without obvious randomization.
  5. The test statistic computed below should reasonably follow a chi-square distribution because all cells in the expected table (Table 1) have values greater than five.
  6. Table 1: Expected frequency table for distribution of hurricanes into the strength and time period categories.

              Cat 1&2 Cat 3 Cat 4&5
    1901-1950    51.4  25.4     8.1
    1951-2000    43.6  21.6     6.9
  7. The table of observed frequencies were given as in Table 2.
  8. Table 2: Observed frequency table for distribution of hurricanes into the strength and time period categories.

              Cat 1&2 Cat 3 Cat 4&5 Sum
    1901-1950      51    26       8  85
    1951-2000      44    21       7  72
    Sum            95    47      15 157
  9. The χ2 test statistic is \(\frac{(51-51.4)^{2}}{51.4}\) + \(\frac{(26-25.4)^{2}}{25.4}\) +\(\frac{(8-8.1)^{2}}{8.1}\) +\(\frac{(44-43.6)^{2}}{43.6}\) +\(\frac{(21-21.6)^{2}}{21.6}\) +\(\frac{(7-6.9)^{2}}{6.9}\) = 0.0036+0.0121+0.0018+0.0043+0.0142+0.0021 = 0.0382 with 2 df.
  10. The p-value is 0.9811.
  11. The H0 is not rejected because the p-value>α.
  12. It appears that the distribution of hurricanes into the strength categories is the same for both time periods (Table 3).
  13. Table 3: Row percentage table for the observations of hurricane strengths by time period.

              Cat 1&2 Cat 3 Cat 4&5 Sum
    1901-1950    60.0  30.6     9.4 100
    1951-2000    61.1  29.2     9.7 100
  14. Not required for a chi-square test.

R Appendix.

library(NCStats)
distrib(0.0382,distrib="chisq",df=2,lower.tail=FALSE)

Response to Hello

  1. α = 0.05.
  2. The H0: “The distribution of groups into whether they responded or not to the ‘Hello’ is the same for all three group sizes” versus H0: “The distribution of groups into whether they responded or not to the ‘Hello’ is NOT the same for all three group sizes.”
  3. A chi-square test is required because (i)a categorical variable with two levels (response or not) was measured on (ii) three groups/populations (group sizes).
  4. This study is experimental in that the the researchers intervened with the subjects (i.e., said “Hello” to them) but it is observational in the sense that the “groups” (number of peple together) were not created by the researcher. It is clear that the “groups” were not randomly selected.
  5. The test statistic below should follow a χ2 distribution because the expected number in each cell is greater than five (Table 4).
  6. Table 4: Expected frequencies of people that responded to ‘Hello’ and group size.

                 Responded Did Not Respond
    Individual        84.3            34.7
    Two or Three      66.6            27.4
    Four or more      19.1             7.9
  7. The observed frequency table is in Table 5.
  8. Table 5: Observed frequencies of people that responded to ‘Hello’ and group size.

                 Responded Did Not Respond Sum
    Individual          92              27 119
    Two or Three        65              29  94
    Four or more        13              14  27
    Sum                170              70 240
  9. The χ2 test statistic is \(\frac{(92-84.3)^{2}}{84.3}\) + \(\frac{(27-34.7)^{2}}{34.7}\) +\(\frac{(65-66.6)^{2}}{66.6}\) +\(\frac{(29-27.4)^{2}}{27.4}\) +\(\frac{(13-19.1)^{2}}{19.1}\) +\(\frac{(14-7.9)^{2}}{7.9}\) = 0.703+ 1.709+ 0.038+ 0.093+ 1.948+ 4.710= 9.201 with (3-1)(2-1)=2 df.
  10. The p-value is 0.0100.
  11. The H0 is rejected because the p-value < α.
  12. There is a difference in the responsiveness among the three sizes of groups. It appears that the groups with 4,5,6 individuals responded less frequently than the individuals and groups of 2 and 3 (Table 6).
  13. Table 6: The percentage of people that responded to ‘Hello’ by group size.

                 Responded Did Not Respond Sum
    Individual        77.3            22.7 100
    Two or Three      69.1            30.9 100
    Four or more      48.1            51.9 100
  14. Not needed for a chi-square test.

R Appendix.

library(NCStats)
distrib(9.201,distrib="chisq",df=2,lower.tail=FALSE)

Turtle Excluder Devices

  • α = 0.10.
  • The H0: “The distribution of tows with zero or at least one turtle mortality is the same between the two sizes of openings” versus HA: “The distribution of tows with zero or at least one turtle mortality is NOT the same between the two sizes of openings.”
  • A chi-square test is required because (i)a categorical variable with two levels (at least one turtle mortality or not) was measured on (ii) two groups/populations (different sized TED openings).
  • An observation study (it seems that the tows were just made and observed, not explicitly controlled by an experimenter) that is not obviously a random sample was used.
  • The test statistic below should follow a χ2 distribution because the expected number in each cell is greater than five (Table 7).
  • Table 7: Expected frequencies of trawl tows with at least one turtle mortality by two sizes of openings.

             mortality no mortality
    original  11.04294     63.95706
    new       12.95706     75.04294
  • The table of observed frequencies is in Table 8.
  • Table 8: Observed frequencies of trawl tows with at least one turtle mortality by two sizes of openings.

             mortality no mortality
    original        16           59
    new              8           80
  • The χ2 test statistic is 4.833 with 1 df.
  • The p-value for this test statistic is 0.0279.
  • The H0 is rejected because the p-value < α.
  • There appears to be a difference in the distribution of tows with zero and at least one turtle mortality between the two opening sizes. In fact, it appears that significantly fewer tows with the new, larger opening had at least one turtle mortality (Table 9).
  • Table 9: The percentage of trawl tows with at least one turtle mortality by two sizes of openings.

             mortality no mortality Sum
    original      21.3         78.7 100
    new            9.1         90.9 100
  • Not needed for a chi-square test.
  • R Appendix

    > obs <- matrix(c(16,75-16,8,88-8),nrow=2,byrow=TRUE)
    > rownames(obs) <- c("original","new")
    > colnames(obs) <- c("mortality","no mortality")
    > ( chi1 <- chisq.test(obs,correct=FALSE) )
    > chi1$expected
    > percTable(chi1$observed,margin=1)