In: Statistics and Probability
An experiment was conducted to determine the effect of a high salt mean on the systolic blood pressure (SBP) of subjects. Blood pressure was determined in 12 subjects before and after ingestion of a test meal containing 10.0 gms of salt. The data obtained were:
Subject |
SBP before meal |
SBP after meal |
1 |
120 |
147 |
2 |
130 |
140 |
3 |
139 |
148 |
4 |
120 |
115 |
5 |
123 |
122 |
6 |
140 |
157 |
7 |
131 |
144 |
8 |
123 |
134 |
9 |
125 |
140 |
10 |
130 |
165 |
11 |
131 |
133 |
12 |
142 |
153 |
Here calculation are done with the help of R software.
(a). Here we can use one sided test also.
(b). Mean of SBP before meal = 129.5
Mean of SBP after meal = 141.5
(c). Standard deviation of SBP before meal = 7.6693
Standard deviation of SBP after meal = 14.2031
(d). Paired t test is appropriate to use on these data.
(e). Hypothesis:
H0: Mean of SBP before meal is equal to mean of SBP after meal.
H1: Mean of SBP before meal is not equal to the mean of SBP after meal.
Test statistic = -3.7358 and critical value = -2.7181
P-value = 0.0032, here p-value is less than level of significance that's why we reject null hypothesis H0.
(f). Yes means are statistically significant.
(g). 99% confidence interval for the difference of two means on SBP = (-12.0412 ,-11.9588)
In the above interval we can see that zero is not included Thus mean difference of two meaans on SBP is not zero.
R code:
########### Paired t-test
######################################
# b = SBP before meal, a = SBP after meal
b = c(120, 130, 139, 120, 123, 140, 131, 123, 125, 130, 131,
142)
a = c(147, 140, 148, 115, 122, 157, 144, 134, 140, 165, 133,
153)
mean(b) ; mean(a)
sd(b); sd(a)
t.test(b,a , paired = TRUE,alternative = "two.sided",
conf.level=0.01)
######## The End
################################################
Output:
> ########### Paired t-test
######################################
> # b = SBP before meal, a = SBP after meal
> b = c(120, 130, 139, 120, 123, 140, 131, 123, 125, 130, 131,
142)
> a = c(147, 140, 148, 115, 122, 157, 144, 134, 140, 165, 133,
153)
> mean(b) ; mean(a)
[1] 129.5
[1] 141.5
> sd(b); sd(a)
[1] 7.669301
[1] 14.20307
> t.test(b,a , paired = TRUE,alternative = "two.sided",
conf.level=0.01)
Paired t-test
data: b and a
t = -3.7358, df = 11, p-value = 0.00329
alternative hypothesis: true difference in means is not equal to
0
1 percent confidence interval:
-12.04118 -11.95882
sample estimates:
mean of the differences
-12
> ######## The End
################################################
>