In: Statistics and Probability
13. A researcher assesses 7 students on test anxiety using blood pressure as a measure (the higher the blood pressure, the greater the anxiety); she then assesses the same subjects again after they view a 2-hour videotape on "relaxation techniques under stress". The results, average systolic blood pressure, were:
Subject Before After
1 120 110
2 160 110
3 124 100
4 135 99
5 170 115
6 143 106
7 188 89
1. State the independent and dependent
variables.
2. State the Null Hypothesis in words and
symbols.
3. Compute the appropriate statistic.
4. What is the decision? reject
5. State the full conclusion in words.
Sample #1 | Sample #2 | difference , Di =sample1-sample2 | (Di - Dbar)² |
120 | 110 | 10.00 | 1185.33 |
160 | 110 | 50.00 | 31.04 |
124 | 100 | 24.00 | 417.33 |
135 | 99 | 36.00 | 71.04 |
170 | 115 | 55.00 | 111.76 |
143 | 106 | 37.00 | 55.18 |
188 | 89 | 99.00 | 2978.04 |
sample 1 | sample 2 | Di | (Di - Dbar)² | |
sum = | 1040 | 729 | 311.000 | 4849.714 |
Ho : µd= 30
Ha : µd ╪ 30
Ho: DIFFERENCE OF BLOOD PRESSURE IS EQUAL BEFORE AND AFTER
RELAXATION TECHNIQUE
Ha: DIFFERENCE OF BLOOD PRESSURE IS not EQUAL BEFORE AND AFTER
RELAXATION TECHNIQUE
Level of Significance , α =
0.05 claim:µd=0
sample size , n = 7
mean of sample 1, x̅1= 148.571
mean of sample 2, x̅2= 104.143
mean of difference , D̅ =ΣDi / n =
44.429
std dev of difference , Sd = √ [ (Di-Dbar)²/(n-1) =
28.4304
std error , SE = Sd / √n = 28.4304 /
√ 7 = 10.7457
t-statistic = (D̅ - µd)/SE = (
44.42857143 - 30 ) /
10.7457 = 1.343
Degree of freedom, DF= n - 1 =
6
p-value =
0.227924 [excel function: =t.dist.2t(t-stat,df) ]
decision p-value>α , Do not reject null
hypothesis
conclusion : DIFFERENCE OF BLOOD PRESSURE IS EQUAL BEFORE AND AFTER RELAXATION TECHNIQUE
Please revert in case of any doubt.
Please upvote. Thanks in advance