In: Math
The following table reports the fasting glucose levels of a sample of potential participants in a research study investigating the efficacy of a new insulin-type drug.
Patient |
Fasting Glucose (mg/dL) |
Patient |
Fasting Glucose (mg/dL) |
A |
117 |
A |
112 |
B |
125 |
B |
132 |
C |
129 |
C |
118 |
D |
116 |
D |
119 |
E |
134 |
E |
134 |
F |
108 |
F |
126 |
G |
127 |
G |
124 |
A) Calculate the mean, median, mode, and standard deviation for
the group.
B) Are there any potential outliers that may be affecting the
statistics calculated in part A?
For Group 1 -
Mean - 122.3
Median - 125
Mode - 108, 116, 117, 125, 127, 129, 134 (All values have only 1 count)
Standard deviation - 8.98
For Group 2 -
Mean - 123.6
Median - 124
Mode - 112, 118, 119, 124, 126, 132, 134 (All values have only 1 count)
Standard deviation - 7.87
B -
For group 1 -
Mean = 122.3
Std = 8.98
Outlier detected? | No |
Significance level: | 0.05 (two-sided) |
Critical value of Z: | 2.01 |
P.S - We can look up the critical vaue of Z at 0.05 significance level form Z table.
Row | Value | Z | Significant Outlier? |
---|---|---|---|
1 | 108. | 1.59 | Furthest from the rest, but not a significant outlier (P > 0.05). |
2 | 116. | 0.70 | |
3 | 117. | 0.59 | |
4 | 125. | 0.30 | |
5 | 127. | 0.53 | |
6 | 129. | 0.75 | |
7 | 134. | 1.31 |
For group 2 -
Mean = 123.6
Std = 7.87
Outlier detected? | No |
Significance level: | 0.05 (two-sided) |
Critical value of Z: | 2.01 |
Row | Value | Z | Significant Outlier? |
---|---|---|---|
1 | 112. | 1.47 | Furthest from the rest, but not a significant outlier (P > 0.05). |
2 | 118. | 0.71 | |
3 | 119. | 0.58 | |
4 | 124. | 0.05 | |
5 | 126. | 0.31 | |
6 | 132. | 1.07 | |
7 | 134. | 1.32 |
P.S - you can look up the p values for calculated Z values at 0.06 significance level from the normal distribution table