Question:
There are various sources of flickering lights in our environment; for instance, light from computer screens and fluorescent bulbs. Whether or not the human eye can detect the flicker depends on its frequency. A 1973 study ("The effect of iris color on critical flicker frequency," Journal of General Psychology [1973], 91- 95) obtained data from a random sample of 19 subjects and recorded eye color and Critical Flicker Frequency (CFF), a numerical measure of threshold sensitivity to flickering light. Do the data suggest that threshold sensitivity to flickering light is related to eye color?
Can we use ANOVA (analysis of variance) to analyze these data?
Yes, because the response variable is quantitative and there are three or more groups of eye colors.
Yes, because ANOVA can be used whenever we have three or more populations being compared regardless of the type of response variable.
No, because even though there are three or more eye colors, the response variable is categorical.
No, because both the explanatory variable and the response variable are quantitative.
Is the equal population standard deviation condition met for this test? Below are the sample standard deviations for each group:
Blue StDev: 1.53 && Green StDev: 1.37 && Brown StDev: 1.84
No, because the largest standard deviation is greater than 2.
Yes, because the largest standard deviation is less than 2.
Yes, because (largest standard deviation) / (smallest standard deviation) is less than 2.
No, because (largest standard deviation) / (smallest standard deviation) is greater than 2
What are the correct null and alternative hypotheses? (In this problem, μBl = mean threshold sensitivity for blue eyes, μG = mean threshold sensitivity for green eyes, and μBr = mean threshold sensitivity for brown eyes)
H0: μBl = μG = μBr vs. Ha: at least one mean is different
H0: μBl = μG = μBr vs. Ha: all of the means are different
H0: μBl = μG = μBr vs. Ha: μBl ≠ μG = μBr
H0: μBl = μG= μBr vs. Ha: μBl ≠ μG ≠ μBr
Assume all the conditions of this test have been met. If the p-value is .010, what can we conclude at α = 0.05?
Reject the null hypothesis and conclude that at least one mean differs from the other means.
Reject the null hypothesis and conclude that there is insufficient evidence to say that the means are not all equal.
Fail to reject the null hypothesis and conclude that there is insufficient evidence to say that the means are not all equal.
Fail to reject the null hypothesis and conclude that at least one mean differs from the other means.
(a) Yes, because the response variable is quantitative and there are three or more groups of eye colors.
(b) Yes, because (largest standard deviation) / (smallest standard deviation) is less than 2.
(c) H₀ : μBl = μG = μBr versus Hₐ : At least one mean is different
(d) Reject the null hypothesis and conclude that at least one mean differs from the other means.
Step-by-step explanation:
(a) ANOVA, which means ANalysis Of Variance consists of statistical models that allow us to analyze the dissimilarity between the means of groups within a sample. ANOVA helps us tests the equality of the means of two or more population.
Since the data consists of recorded eye color and Critical Flicker Frequency (C. F .F.), the response is the C. F. F. which is quantitative, and from which we can get the group mean, the correct option is;
Yes, because the response variable is quantitative and there are three or more groups of eye colors.
(b) The sample standard deviation for each group are;
Blue StDev: 1.53 && Green StDev: 1.37 && Brown StDev: 1.84
It is the convention that ANOVA is appropriate where the ratio of the largest sample standard deviation to the smallest sample standard deviation is less than two. Therefore, the correct option is;
Yes, because (largest standard deviation) / (smallest standard deviation) is less than 2.
(c) Since the ANOVA is to help us test the equality of the mean of the groups of data in a sample then the null hypothesis should be;
H₀ : μBl = μG = μBr versus Hₐ : At least one mean is different
(d) The null hypothesis is rejected if the p-value is less than or equal to alpha and the result is then statistically significant.
Here the p-value is 0.01 and α = 0.05, therefore our p-value is statistically significant at the α level of 0.05 and we reject the null hypothesis. That is;
Reject the null hypothesis and conclude that at least one mean differs from the other means.
