May someone please review my findings.
Also, is my part D correct?
Thank you in advance.
Starting Weight Final Weight Weight Loss A company wants to (cheaply) test the effects of a weight loss drug they're developing. They claim that the 282 282 Determine for weight loss column drug will help arry person who is overweight lose $ pounds in a week. They decide to conduct a hypothesis 15 5 Sample Mean: 4.425 test at the 99% significance level to further strengthen their claim. Thus they decide to compensate a group 280 27815 188 17915 S Null Population Mean: of 40 people who are overweight to take this drug for a week and come back for a weight in. 273 267 Sample Standard Deviation: 3.378950679 201 194 7 Sample Size: They decided to proceed with a hypothesis test and they want to disprove the null hypothesis that their drug 180 171 9 Standard Error: 0.534259012 does not help people who are overweight with weight loss (that is, m=0) 302.5 3.128887636 308 Cl Lower Limit Part A: State the null hypothesis and alternative hypotheses for this particular test. 219 21315 Ci Upper Limit 5.721112364 196 193 3 Null hypothesis : The weight loss is less than or equal to 5lbs. WL-5 50 181 180 1 Perform Hypothesis Test Ho : H = 0 HO : UI = D 195 187 T-Statistic: -1.076256996 Alternative hypothesis: The drug will not help overweight people lose 5lbs in one week. WL-5 >0 248 244 T-critical Value(s) -2.426, 2.426 H1 : [>0 NO 276 275.5 D.5 Reject the Null Hypothesis? 213 209 Part B: Use the data given to determine the following statistical measures and find the confidence interval at 204 20315 99% for the average weight loss in one week. Is a 5 pounds loss contained within this confidence interval? if 270 270 0 so, does that mean that customers can expect to lose 5 pounds on this drug? 303 293.5 263 257 6 T-Critical Value excel formula 2.423256779 Cl = X + z s+vn 294 293 = (3.128887636, 5721112364) 277 270 7 191 189.5 1.5 I already obtained the confidence interval in the using the excel formula in the table to left 184 175 9 which is 3.128887636 - 5721112364. The number 5 is contained in this interval. This 245 242.5 2.5 means that it is a possibility that the customers can lose 5 Ibs in one week. This is only a 198 196.5 1.5 possibility. More testing needs to be done for accurate results. 296 287 4 194.5 1.5 Part C: Perform a hypothesis test! Which type of tailed-test will you use? Find the t-statistic for the data and 196 determine whether to reject the null hypothesis. Hint: the critical value of the t-distribution with 39 degrees 296 295 of freedom is approximately 2.426. If you have two critical values, separate them with commas and list your 239 229 positive value first in the T-Critical Value(s) cell. 286 285 232 226 the t-statistic is equal to -1.07626. The critical values are 2.426 and -2.426. The rejection 190 190 region is above 2.426. If the t-statistic falls in that region, I will reject the null hypothesis. 218 217.5 280 272 314 304.5 9.5 Part D: Were you able to reject the null hypothesis at 99% significance? If so, interpret the significance of 227 223 this rejection. Does it provide sufficient evidence to prove that their drug helps their customers lose 5 285 279 pounds in one week? If you think this is insufficient evidence, describe how you would change the 191 191 experiment to make it more meaningful. 188 181 298 290 Since the t-statistic is less than the critical value, (-1.076
