Question

total, 16 patients were enrolled in the study. Values of the f-wave frequency during day- and night-time are given in the table below.

Time of the day	Aggregation values
Day-time (n = 16)	6.23	6.91	6.35	6.29	6.45	6.30	6.60	6.54	6.64	6.90	6.11	7.28	6.93	7.89	7.21	6.90
Night-time (n = 16)	6.41	5.98	6.25	6.03	6.57	6.25	6.51	6.50	6.50	6.41	6.70	6.03	6.60	6.77	6.88	6.93

Data analysis.

1. Summarize the data.

(decide which sampling technique was used to collect the data;

check if the data is normally distributed and if the variances of the groups are similar;

present data graphically; briefly interpret the results)

Answer 1

Sampling technique: The data is probably an average for multiple readings (say 4 or 5 readings) for each of the 16 patients during day time and during night time. Taking multiple readings would reduce errors and get a closer to true value for each patient.

Below is the data summary:

Time of the day	Average	Std Dev	Median	Minimum	Maximum
Day-time (n = 16)	6.720625	0.469332	6.62	6.11	7.89
Night-time (n = 16)	6.4575	0.292313	6.5	5.98	6.93

Both mean and median are greater for Day-time and less for Night-time data. The Minimum and Maximum are also greater. In 12 out of 16 patients, that is 75% cases, the Day-time frequency is greater than Night-time frequency.

Below is the histogram and QQ-plots vs. normal distribution. The distribution are close but not normal. The first is skewed to right and second shows slight left skew with a much higher mode.

Comparison of Variances:

Standard Error for Day-Time = 0.469332

Standard Error for Night-time = 0.292313

Standard Error is an estimate for square root of Variance.

Therefore Day-time has greater variance. Test of Hypothesis for equality of two variances is the F-test.

The Test Statistic is

F = Var(1) / Var(2)

where Var(1) and Var(2) are the sample variances in groups 1 and 2.

The degrees of freedom for F are N_1 - 1 and N_2-1 where N_1 and N_2 are sample sizes of groups 1 and 2.

Null Hypothesis is that Day-time variance = Night-time variance

Alternative Hypothesis for 1-tailed is that Day-time variance is greater.

F statistic for the given data is

F = (0.469332*0.469332) / (0.292313*0.292313)

F (df1 = 15, df2 = 15) = 2.577899

With alpha = 0.05 for a 1-tailed distribution

check if F is greater than the critical value.

F_critical_upper_limit = 2.40

Reject null hypothesis of equality of variances if F > 2.40

F-value = 2.577899

We can reject the null hypothesis at significance level alpha = 0.05 and accept alternative hypothesis that the Day-Time observations show greater variance. This is also clear from the histograms, because the Day-time data shows greater spread (range for day-time = 1.78 vs. range for night-time = 0.95), whereas Night-time shows a sharp mode and is centered around the mode.