Question

The medical researcher is comparing two treatments for lowering cholesterol: diet and meds. The researcher wants to see if the patients who receive the recommendation to change their diet have equal success lowering cholesterol compared to a prescription of meds. A random sample of some patients who received the recommendation to change their diet and others who were prescribed meds was taken. The results of how many did or did not lower their cholesterol are shown below: Data on Diet vs. Meds for Weight Loss Diet Meds Yes 450 458 No 233 263 What can be concluded at the alpha = 0.01 level of significance? For this study, we should use The null and alternative hypotheses would be: H_0: (please enter a decimal) H_1: (Please enter a decimal) The test statistic = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is alpha Based on this, we should the null hypothesis. Thus, the final conclusion is that ... The results are statistically significant at α = 0.01, so there is sufficient evidence to conclude that the success rate for the 683 patients who received the recommendation to change their diet is different from the success rate for the 721 patients who were prescribed meds. The results are statistically significant at α = 0.01, so there is sufficient evidence to conclude that the population of all patients who received the recommendation to change their diet is not equally likely to lower their cholesterol as the population of patients who are prescribed meds. The results are statistically insignificant at α = 0.01, so there is insufficient evidence to conclude that the population of all patients who received the recommendation to change their diet is not equally likely to lower their cholesterol as the population of patients who are prescribed meds. The results are statistically insignificant at α = 0.01, so we can conclude that the success rate for all patients who receive the recommendation to change their diet is equal to the success rate for all patients who are prescribed meds. Interpret the p-value in the context of the study. If the success rate for the sample of patients who receive the recommendation to change their diet is the same as the success rate for the sample of patients who were prescribed meds and if another 683 patients are given the recommendation to change their diet and 721 patients are prescribed meds then there would be a 35.46% chance of concluding that the difference in the success rate for all patients who receive the recommendation to change their diet and all patients who are prescribed meds is at least 2.4%. If the success rate for the population of patients who receive the recommendation to change their diet is the same as the success rate for the population of patients who are prescribed meds and if another 683 patients who are given the recommendation to change their diet and 721 patients who are prescribed meds are surveyed then there would be a 35.46% chance that the percent of the surveyed diet changers who lowered their cholesterol would differ at least 2.4% compared to the percent of the surveyed med takers who lowered their cholesterol. There is a 35.46% chance of a Type I error. There is a 35.46% chance that the difference in the success rate for all patients who receive the change of diet recommendation and all patients who are prescribed meds is at least 2.4%. Interpret the level of significance in the context of the study. There is a 1% chance that a patient won't be able to afford the meds, so they might as well change their diet. If the success rate for the population of patients who receive the recommendation to change their diet is the same as the success rate for the population of patients who are prescribed meds and if another 683 patients who are given the recommendation to change their diet and 721 patients who are prescribed meds are surveyed then there would be a 1% chance that we would end up falsely concuding that the proportion of the 683 patients who received the diet change recommendation who lowered their cholesterol is different from the proportion of the 721 patients who were prescribed meds who lowered their cholesterol. If the success rate for the population of patients who receive the recommendation to change their diet is the same as the success rate for the population of patients who are prescribed meds and if another 683 patients who are given the recommendation to change their diet and 721 patients who are prescribed meds are surveyed then there would be a 1% chance that we would end up falsely concuding that the success rate for the population of patients who receive the recommendation to change their diet is different from the success rate for the population of patients who are prescribed meds.

Answer 1

Here , we are asked to test whether the patients who receive the recommendation to change their diet have equal success lowering cholesterol compared to a prescription of meds.In order to test the above statement, we may compare two treatments for lowering cholesterol: diet and meds.

Here, we are given count data on the subjects under a diet regime or meds, who succeded in lowering their cholesterol levels. Here, the best measure for comparison would be 'Proportion'.

Let

X1 = Event that a person under a diet succeded in lowering his cholesterol levels X2 = Event that a person under meds succeded in lowering his cholesterol levels

By definition of proportion, the sample proportions may be computed as:

Here, at the alpha = 0.01 level of significance, we may state there is a 99% (1-.01=.99) chance of the test result being true.We may compare the p-value against the alpha to decide whether to reject the null hypothesis or not. If the p-value (Probability of obtaining an extreme result, given the null is true) is less than alpha, we would have sufficient evidence to reject the null.

To test:

Vs   ...........................................................(1)

The appropriate statistical test to test the above hypothesis would be a Two sample Z test for Proportion.

Computing the test statistic, given by the formula,

where,

Substituting the given values,

= 0.926  ..........................................................(2)

To obtain the p-value of the test,

From Standard Normal table:

For two tailed test,

P-value = 2( 1 - 0.82381)

= 0.3545  ..........................................................(3)

The p-value = 0.3545 >  0.01. Based on this, we fail to Reject the null hypothesis.

Thus, the final conclusion is that:

The results are statistically insignificant at α = 0.01, so there is insufficient evidence to conclude that the population of all patients who received the recommendation to change their diet is not equally likely to lower their cholesterol as the population of patients who are prescribed meds.

Here, the conclusions, p-value and alpha are all tools of Hypothesis testing for drawing inferences on the population based on the sample. Hence, all conclusions would be on the population, in general and not on the sample alone.

Interpreting the p-value in the context of the study:

If the success rate for the population of patients who receive the recommendation to change their diet is the same as the success rate for the population of patients who are prescribed meds and if another 683 patients who are given the recommendation to change their diet and 721 patients who are prescribed meds are surveyed then there would be a 35.46% chance that the percent of the surveyed diet changers who lowered their cholesterol would differ at least 2.4% compared to the percent of the surveyed med takers who lowered their cholesterol.

Interpreting the level of significance in the context of the study:

If the success rate for the population of patients who receive the recommendation to change their diet is the same as the success rate for the population of patients who are prescribed meds and if another 683 patients who are given the recommendation to change their diet and 721 patients who are prescribed meds are surveyed then there would be a 1% chance that we would end up falsely concuding that the success rate for the population of patients who receive the recommendation to change their diet is different from the success rate for the population of patients who are prescribed meds.