(5 points) How, if at all, does caffeine intake affect hormone levels in women?
ID: 3360479 • Letter: #
Question
(5 points) How, if at all, does caffeine intake affect hormone levels in women? Lucero et al. (2001) report a study involving a sample of 498 women from the general population aged between 36 and 45. The women each completed a questionnaire, including items on caffeine intake, and provided a blood sample collected during the early phase of the menstrual cycle. We focus here on the level of estradiol (E2), and how that may depend on daily caffeine intake (in mg per day). Suppose the scatter plot of the logarithm of the estradiol level (Y) against daily caffeine intake (X) appeared as below: Log Estradiol Level Against Daily Caffeine Intake Cafeine intake (mgiday) A linear regression model fitted to the above scatter plot is A linear regression model fitted to the above scatter plot is Y = 3.338 0.000608 Jenniter Lucero, Bemard L. Harlow, Robert L. Barbieri, Patrick Sluss, Daniel W. Cramer (2001): Earty follicular phase hormone levels in relation to patterns of alcohol, tobacco, and coffee use. Fertility and Steriity 76, Issue 4, 723-729. Part a) Suppose one of the subjects reported a daily caffeine intake of 155.99 mg/day and had a recorded log(E2) level of 3.44. Compute the residual for this observation, to two decimal places. Part b) Suppose the ANOVA table for the regression is as below Source DoF SS MS F 6.54 () 6.298 ii) 12.838 Caffeine Error Total Provide the missing entries, to two decimal places. Part c) Based on the above information we wish to test the null hypothesis that log(E2) level does not depend linearly on daily caffeine intake Provide the test statistic that would appear in the ANOVA table, to two decimal places. Part d lf testing at the 1% significance level, would you reject or not reject the null hypothesis that log dally caffeine intake? 2 level does not depend linearly on A. Not reject B. RejectExplanation / Answer
a) Given
Predicted vlaue : Y=3.338 +0.000608(155.99)
=3.4328
Actual value = 3.44
Answer: Residual : Actual value - Predicted value = 3.44 - 3.4328 = 0.0072 = 0.01
b) From the given data
(i) 6.54/1 = 6.54
(ii) 6.298 / 496 = 0.012697 = 0.01
c)
F-statistic = 6.54/0.01 = 515.058
Correct Answer : Not rejected
Source df Sum of square mean Square F P-value Regression 1 6.54 6.54 515.058749 9.8745E-79 Error 496 6.298 0.012697581 Total 497 12.838Related Questions
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