In: Statistics and Probability
3. Calculate the p-value (3dp) for a t-test on the chol.csv data with the correct null hypothesis from the lab test exercise.
4.Using Fishers F-test determine whether you can accept or reject the null hypothesis that there is no difference between the variances of the day_2 and day_4 samples.
chol.csv
| day_2 | day_4 | 
| 270 | 218 | 
| 236 | 234 | 
| 210 | 214 | 
| 142 | 116 | 
| 280 | 200 | 
| 272 | 276 | 
| 160 | 146 | 
| 220 | 182 | 
| 226 | 238 | 
| 242 | 288 | 
| 186 | 190 | 
| 266 | 236 | 
| 206 | 244 | 
| 318 | 258 | 
| 294 | 240 | 
| 282 | 294 | 
| 234 | 220 | 
| 224 | 200 | 
| 276 | 220 | 
| 282 | 186 | 
| 360 | 352 | 
| 310 | 202 | 
| 280 | 218 | 
| 278 | 248 | 
| 288 | 278 | 
| 288 | 248 | 
| 244 | 270 | 
| 236 | 242 | 
3. H0: There is no significance difference between the means of day_2 and day_4
H1: There is significance difference between the means of day_2 and day_4
Let the los be alpha = 5%
From the given data


Critical t: ±2.004877
P-Value: 0.0712
Degrees of freedom: 54.0000
Here t value is lies between t critical value and p-value > alpha 0.05 so we accept H0
Thus we conclude that There is no significance difference between the means of day_2 and day_4
4. H0: There is no significance difference between the variance of day_2 and day_4
H1: There is significance difference between the variance of day_2 and day_4
Let the los be alpha = 5%

Test Statistic, F: 1.0319
Lower Critical F: 0.4627611
Upper Critical F: 2.160946
P-Value: 0.9356
Here F value is lies between F critical values and P-value is > alpha 0.05 so we accept H0
Thus we conclude that there is no significance difference between the variance of day_2 and day_4