In: Statistics and Probability
The table below shows the weights of seven subjects before and after following a particular diet for two months. Subject / A / B / C / D / E / F / G ////////////////////////////////////////////////// Before / 165 / 151 / 196 / 175 / 179 / 172 / 191 ////////////////////////////////////////////////// After / 166 / 154 / 169 / 164 / 176 / 193 / 169 Using a 0.01 level of significance, test the claim that the diet is effective in reducing weight. Use the critical value method of hypothesis testing.
paired t test
SAMPLE 1 | SAMPLE 2 | difference , Di =sample1-sample2 | (Di - Dbar)² |
165 | 166 | -1 | 41.327 |
151 | 154 | -3 | 71.041 |
196 | 169 | 27 | 465.327 |
175 | 164 | 11 | 31.041 |
179 | 176 | 3 | 5.898 |
172 | 193 | -21 | 698.469 |
191 | 169 | 22 | 274.612 |
Ho : µd= 0
Ha : µd > 0
mean of difference , D̅ =ΣDi / n =
5.429
std dev of difference , Sd = √ [ (Di-Dbar)²/(n-1) =
16.267
std error , SE = Sd / √n = 16.2671 /
√ 7 = 6.1484
t-statistic = (D̅ - µd)/SE = (
5.428571429 - 0 ) /
6.1484 = 0.8829
Degree of freedom, DF= n - 1 =
6
α=0.05
t-critical value , t* =
1.9432 [excel function: =t.inv(α,df) ]
Decision: test stat < 1.9432 , Do not reject null
hypothesis
there is no enough evidence to conclude that diet is effective in reducing weight
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