In: Statistics and Probability
A medical researcher is interested in whether calcium intake differs, on average, among elderly individuals with normal bone density, with osteopenia (a low bone density that may lead to osteoporosis), and with osteoporosis. He recruits consenting 60 individuals over the age of 65 using his patient records from his hospital (20 from each of the three bone density groups). You may assume that the individuals in each group constitute a simple random sample from consenting patients from that hospital. The individuals keep a food diary for two weeks, at the end of which the researcher computes their average daily calcium intake (in mg). The data are available in the file calcium.csv.
calcium.csv file=
"calcium","group(normal or osteoporosis)"
861,"normal" 884,"normal" 1009,"normal" 905,"normal" 909,"normal" 1020,"normal" 932,"normal" 811,"normal" 852,"normal" 869,"normal" 986,"normal" 925,"normal" 928,"normal" 908,"normal" 861,"normal" 1025,"normal" 935,"normal" 762,"normal" 949,"normal" 867,"normal" 675,"osteopenia" 735,"osteopenia" 678,"osteopenia" 699,"osteopenia" 706,"osteopenia" 632,"osteopenia" 809,"osteopenia" 761,"osteopenia" 670,"osteopenia" 838,"osteopenia" 780,"osteopenia" 729,"osteopenia" 813,"osteopenia" 811,"osteopenia" 808,"osteopenia" 798,"osteopenia" 789,"osteopenia" 746,"osteopenia" 729,"osteopenia" 723,"osteopenia" 651,"osteoporosis" 685,"osteoporosis" 611,"osteoporosis" 852,"osteoporosis" 785,"osteoporosis" 621,"osteoporosis" 672,"osteoporosis" 667,"osteoporosis" 755,"osteoporosis" 694,"osteoporosis" 718,"osteoporosis" 698,"osteoporosis" 697,"osteoporosis" 796,"osteoporosis" 684,"osteoporosis" 806,"osteoporosis" 592,"osteoporosis" 741,"osteoporosis" 709,"osteoporosis" 715,"osteoporosis"
a. What are the null and alternative hypotheses?
b. (1 mark) What is the value of the test statistic?
c. (1 mark) What is the p-value?
d. Using a significance level of = 0.05, state your conclusions in the language of the problem.
e. State and verify (using plots and/or descriptive statistics) the additional two assumptions required for the p-value in c) to be valid.
b. Value of test statistic=F=55.2110
Anova: Single Factor | ||||||
SUMMARY | ||||||
Groups | Count | Sum | Average | Variance | ||
Normal | 20 | 18198 | 909.9 | 4641.463 | ||
Osteopenia | 20 | 14929 | 746.45 | 3389.208 | ||
Osteoporosis | 20 | 14149 | 707.45 | 4507.208 | ||
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Between Groups | 461486 | 2 | 230743.017 | 55.21102 | 4.61E-14 | 3.158843 |
Within Groups | 238219.7 | 57 | 4179.29298 | |||
Total | 699705.7 | 59 |
c. P-value=4.61x 10-14.
d. Since P-value<0.05 so we reject null hypothesis at 5% level of significance and conclude that there is sufficient evidence that calcium intake differs significantly, on average, among elderly individuals with normal bone density, with osteopenia and with osteoporosis.
e. Here we assume normality and equal variances.
From above plot we observe that normality assumption holds.
Test for Equal Variances: Daily calcium intake versus Group
95% Bonferroni confidence intervals for standard deviations
Group N Lower StDev Upper
Normal 20 48.9334 68.1283 109.072
Osteopenia 20 41.8145 58.2169 93.204
Osteoporosis 20 48.2205 67.1357 107.483
Bartlett's Test (Normal Distribution)
Test statistic = 0.54, p-value = 0.763
Levene's Test (Any Continuous Distribution)
Test statistic = 0.03, p-value = 0.970
From above tests we see that p-value>0.05 so assumption of equal variances is also true.