- Description of the distribution of calcium intake.
The distribution is normal with a tail towards the right. The distribution is also less kurtic since it does not have a sharp top. It is also characterized by a gap in between the bars.
- Descriptive statistics
Column1
|
Mean
|
941.8157895
|
Standard Error
|
72.77628824
|
Median
|
845
|
Mode
|
#N/A
|
Standard Deviation
|
448.6231703
|
Sample Variance
|
201262.7489
|
Kurtosis
|
2.323888221
|
Skewness
|
1.333427894
|
Range
|
2059
|
Minimum
|
374
|
Maximum
|
2433
|
Sum
|
35789
|
Count
|
38
|
Table 1 (generated from excel)
- 95% C.I for the mean consumption of calcium.
C.I is calculated as below,
The formula is = Mean = 941.81578
n = 38
Z value for 95% from normal tables = 1.65
941.82
- Mean and median as a percentage of RDA.
If RDA = 1200
Mean = 941.82
Median = 845
Mean as a percentage of RDA =
Mean as a percentage of RDA =
- Transformed data descriptive statistics.
Column1
|
Mean
|
78.48464912
|
Standard Error
|
6.064690687
|
Median
|
70.41666667
|
Mode
|
#N/A
|
Standard Deviation
|
37.38526419
|
Sample Variance
|
1397.657979
|
Kurtosis
|
2.323888221
|
Skewness
|
1.333427894
|
Range
|
171.5833333
|
Minimum
|
31.16666667
|
Maximum
|
202.75
|
Sum
|
2982.416667
|
Count
|
38
|
Table 2
C.I is calculated as below,
The formula is = Mean = 78.484
n = 38
Z value for 95% from normal tables = 1.65 78.484
- Since we divided the intake by 1200 and multiplied by 100, we can verify that we can obtain the same results by performing the inverse. That is, we multiply by 1200 then divide by 100.
Taking the mean,
Same to the interval value,
95% confidence for the mean of blood sodium concentration
Column1
|
Mean
|
110.9
|
Standard Error
|
3.518298034
|
Median
|
110.5
|
Mode
|
95
|
Standard Deviation
|
15.73430714
|
Sample Variance
|
247.5684211
|
Minimum
|
87
|
Maximum
|
152
|
Sum
|
2218
|
Count
|
20
|
Table 3
The formula is Mean = 110.9
n = 20
Z value for 95% from normal tables = 1.65 110.9
- Is the treatment effective after 12 months?
To establish the effect, we perform a t-test of before and after to see whether the change was significant.
Our null and alternative hypothesis is as below;
H0: There is no significant difference in the blood sodium levels before and after the treatment.
Versus
H1: There is a significant difference in the blood sodium levels before and after the treatment.
Level of significance is 0.05
Paired Samples Test
|
|
Paired Differences
|
t
|
df
|
Sig. (2-tailed)
|
Mean
|
Std. Deviation
|
Std. Error Mean
|
95% Confidence Interval of the Difference
|
Lower
|
Upper
|
before - after
|
-12.30000
|
14.55697
|
3.25504
|
-19.11287
|
-5.48713
|
-3.779
|
19
|
.001
|
Table 4
So as to arrive at the conclusion, we should compare the p-value and significance level.
In this case the p-value is less than the level of significance (.001<.05), therefore we reject the null hypothesis and accept the alternative. So we conclude that there is a significant difference in the blood sodium levels before and after the treatment.
References
Richler, J. (2012). Behaviour research methods.
Woodward, W. (2007). Comparing two means using t-test.