90 likes | 288 Views
Chi squared . A biologist randomly selects 20 fruit flies. All of them are placed in a tube with vinegar on one side and mustard on the other side. Do these fruit flies have a preference for either mustard or vinegar? The results are listed below. mustard : 5 vinegar: 15 .
E N D
A biologist randomly selects 20 fruit flies. All of them are placed in a tube with vinegar on one side and mustard on the other side. Do these fruit flies have a preference for either mustard or vinegar? The results are listed below. mustard : 5 vinegar: 15 First name the test. χ2 Goodness of Fit Test.
A biologist randomly selects 20 fruit flies. All of them are placed in a tube with vinegar on one side and mustard on the other side. Do these fruit flies have a preference for either mustard or vinegar? The results are listed below. mustard : 5 vinegar: 15 χ2 Goodness of Fit Test. Write the hypotheses Ho : Fruit flies are equally drawn to mustard and vinegar. Ha : Fruit flies are not equally drawn to mustard and vinegar.
A biologist randomly selects 20 fruit flies. All of them are placed in a tube with vinegar on one side and mustard on the other side. Do these fruit flies have a preference for either mustard or vinegar? The results are listed below. mustard : 5 vinegar: 15 χ2 Goodness of Fit Test. Ho : The fruit flies are equally drawn to mustard and vinegar. Ha : The fruit flies are not equally drawn to mustard and vinegar. Conditions The fruit flies are randomly selected is given. Data are categorical (or counts). All expected values are ≥ 5. Expected counts for each group = 10
Calculate χ2 Degrees of freedom = 2 – 1 = 1
χ2 = 5 df= 1 .025 < p-value < .05
Biology table for chi-squared critical value approach So how much proof do you need?
χ2 Goodness of Fit Test. Ho : The fruit flies are equally drawn to mustard and vinegar. Ha : The fruit flies are not equally drawn to mustard and vinegar. Conditions The fruit flies are randomly selected is given. Data are categorical (or counts). All expected values are ≥ 5. Expected counts for each group = 10 χ2 = 5 > 3.84 df = 1 .025 < p-value < .05 Conclusion Reject Ho since the p-value < α = .05 (or since χ2 > critical value 3.84) We have evidence that fruit flies do have a preference. They appear to prefer vinegar over mustard. Ta Da!