Question

3. [20 marks] A double-blind randomised experiment investigated whether consumption of oligofructose would stimulate calcium absorption in healthy human males aged 14 to 16 years. Two sets of pills were prepared, identical in appearance and packaging. One set contained oligofructose and the other set contained ordinary sucrose (i.e., a placebo). The experiment ran in two rounds, with the second round starting 3 weeks after the first ended. Subjects were randomly assigned to either receive oligofructose in the first round and the placebo in the second, or the placebo in the first round and oligofructose in the second. In each round, subjects took pills for nine days and had their calcium absorption measured on the last day. Calcium absorption was measured as a percentage of calcium intake. The 3 week period between rounds ensured that the sugars (oligofructose and sucrose) had been completely excreted before the second round commenced.

The calcium absorption data are available in two data files, Oligo_paired.csv and Oligo_indep.csv. Only one of these is appropriate to the problem, and the same calcium absorption measurements appear in each file.

In Oligo_paired.csv, the data columns are headed “Placebo” and “Oligofructose” to indicate which pill was being taken when calcium absorption was measured. Each row contains two calcium absorption measurements from a single subject.

In Oligo_indep.csv, the columns are headed “CaAbsorb” and “Pill”. CaAbsorb contains calcium absorption measurements, and “Pill” indicates whether the pill being taken before each measurement was a placebo or oligofructose.

(a) [Note: this part is worth 4 marks because it is important to the rest of the question, not because it needs a long answer. A correct answer can be quite short. Follow-through marks will be available if you get this part wrong.] Should the data be analysed as independent samples, or as paired data? Justify your answer. [4]

(b) Provide a graphical display of the data, consistent with your choice of analysis. [4]

(c) Formally test whether oligofructose stimulates calcium absorption in healthy males aged 14 to 16 years. You do not need to revisit the decision made in part (a), but your report should be consistent with that decision, and should include all relevant elements of a hypothesis test. Assume α = .05

1 CaAbsorb Pill 78.4 Placebo 76.6 Placebo 57.4 Placebo 51.5 Placebo 49 Placebo 46.6 Placebo 44.2 Placebo 42.9 Placebo 37.2 Placebo 34.1 Placebo 24.6 Placebo 10 12 62 Oligofructose 95.1 Oligofructose 46.5 Oligofructose 49.4 Oligofructose 89.7 Oligofructose 43.8 Oligofructose 50.3 Oligofructose 51.6 Oligofructose 66.6 Oligofructose 52.7 Oligofructose 54 Oligofructose 15 16 17 18 19 20 21 23 Oligo indep Placebo ecovery 1 Placebo Oligofructose 2 e following files. Save 3 4 78.4 62 76.6 95.1 46.5 49.4 89.7 46.643.8 44.250.3 51.6 66.6 34.1 52.7 54 eep 57.4 51.5 49 version 1).xlsb [A... ted from the last. 12:21 AM 7 sv [Originall ated last time the... 5:01 PM 42.9 37.2 10 12 13 24.6 14 Oligo paired 15

Answer 1

(a) this is paired data as the SAME participants are measured at different time intervals and compared. The paired sample t-test, sometimes called the dependent sample t-test, is a statistical procedure used to determine whether the mean difference between two sets of observations is zero. In a paired sample t-test, each subject or entity is measured twice, resulting in pairs of observations.

(b) please find the box-plot of the given data, it may appropriate to compare the two treatments

1:placebo , 2:oligofructose

2 8 06 08 04 09 OSOFOC

(c) null hypothesis H0:$\bar{d}$=0 ( there is no stimulation)

alternate hypothesis H0:$\bar{d}$<0 ( there is stimulation), this is one tailed test

d=pacebo-Oligofructose

here we use t-test and one tailed p-value 0.0379 is less than typical alpha=0.05, so we fail to accept null hypothesis and conclude that there is stimulation and oligofructose stimulates calcium absorption in healthy males aged 14
to 16 years..

s.n.	placebo	Oligofructose	d=pacebo-Oligofructose
1	78.4	62	16.4
2	76.6	95.1	-18.5
3	57.4	46.5	10.9
4	51.5	49.4	2.1
5	49	89.7	-40.7
6	46.6	43.8	2.8
7	44.2	50.3	-6.1
8	42.9	51.6	-8.7
9	37.2	66.6	-29.4
10	34.1	52.7	-18.6
11	24.6	54	-29.4
		n=	11
		mean=d_bar	-10.8364
		sd=	18.1512
		SE(d_bar)=sd/sqrt(n)=	5.4728
		t=d_bar/SE(d_bar)=	-1.9800
		critical t=	2.2281
	one tailed	p-value=	0.0379