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