In: Statistics and Probability
Ten individuals have their systolic blood pressures measured while they are in the dentist’s waiting room and again an hour after the conclusion of the visit to the dentist. The data are as follow: (5 Points).
B.P before |
132 |
135 |
149 |
133 |
119 |
121 |
128 |
132 |
119 |
110 |
B.P after |
118 |
137 |
140 |
139 |
107 |
116 |
122 |
124 |
115 |
103 |
Write a four step procedure to conduct test of hypothesis for the above data
Test the hypothesis at 3% level of significance whether mean difference of the blood pressure before and after visit is zero or not.
Answer:
Given that,
Ten individuals have their systolic blood pressures measured while they are in the dentist’s waiting room and again an hour after the conclusion of the visit to the dentist.
The data are as follow:
B.P before | 132 | 135 | 149 | 133 | 119 | 121 | 128 | 132 | 119 | 110 |
B.P after | 118 | 137 | 140 | 139 | 107 | 116 | 122 | 124 | 115 | 103 |
The given data for n = 10 individual ara a Paired data set so we use paired t-test for the Inference for a Difference in Means with Paired Data.
The null and alternative hypothesis are:
i.e., the mean of the difference of values of blood pressure Before and After visit is NOT DIFFERENT than zero.
i.e., the mean of the difference of values of blood pressure Before and After visit is DIFFERENT than zero.
At significance level of 3%, i.e., =0.03 we need to test the hypothesis.
The test-statistic:
The formula for the test statistic is given by,
and it follows a t-distribution with degrees of freedom, .
where,
: number of matched pairs, i.e., =10
: sample mean difference between matched pairs.
: sample standard deviation of differences of matched pairs.
B.P Before | B.P After | d=Before- After |
132 | 118 | 14 |
135 | 137 | -2 |
149 | 140 | 9 |
133 | 139 | -6 |
119 | 107 | 12 |
121 | 116 | 5 |
128 | 122 | 6 |
132 | 124 | 8 |
119 | 115 | 4 |
110 | 103 | 7 |
Given:
=10
Calculation of test-statistic:
So the test statistic is calculated as .
P-value:
Since we are testing a two-tailed hypothesis and the test statistic is calculated as t=2.99448 then the p-value is given by-
So the P-value is calculated as .
Decision:
The p-value is P-value=0.0151 and the signifiance level is =0.03.
At significance level of =0.03 the sample data provides sufficient evidence to reject null hypothesis H0 .
Hence, we accept the alternative hypothesis Ha ,
i.e.,
In other words, since we rejected the null hypothesis, so we conclude that Before and After visit Blood pressure is NOT EQUAL its different.