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 more 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	492	495
No	189	216

What can be concluded at the  = 0.05 level of significance?

For this study, we should use Select an answer t-test for a population mean z-test for a population proportion z-test for the difference between two population proportions t-test for the difference between two independent population means t-test for the difference between two dependent population means

1. The null and alternative hypotheses would be:   
2.   

   Select an answer p1 μ1  Select an answer ≠ = > <  Select an answer μ2 p2  (please enter a decimal)   

   Select an answer μ1 p1  Select an answer > ≠ < =  Select an answer μ2 p2  (Please enter a decimal)

2. The test statistic ? t z  =  (please show your answer to 3 decimal places.)
3. The p-value =  (Please show your answer to 4 decimal places.)
4. The p-value is ? ≤ >  
5. Based on this, we should Select an answer reject fail to reject accept  the null hypothesis.
6. Thus, the final conclusion is that ...
   • The results are statistically significant at αα = 0.05, so there is sufficient evidence to conclude that the success rate for the 681 patients who received the recommendation to change their diet is greater than the success rate for the 711 patients who were prescribed meds.
   • The results are statistically insignificant at αα = 0.05, so there is insufficient evidence to conclude that the population of all patients who received the recommendation to change their diet is more likely to lower their cholesterol than the population of patients who are prescribed meds.
   • The results are statistically insignificant at αα = 0.05, 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.
   • The results are statistically significant at αα = 0.05, so there is sufficient evidence to conclude that the population of all patients who received the recommendation to change their diet is more likely to lower their cholesterol than the population of patients who are prescribed meds.
7. Interpret the p-value in the context of the study.
   • There is a 14.04% chance of a Type I error.
   • 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 681 patients are given the recommendation to change their diet and 711 patients are prescribed meds then there would be a 14.04% chance of concluding that the success rate for all patients who receive the recommendation to change their diet is at least 2.6% larger than the success rate for patients who are prescribed meds.
   • 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 681 patients who are given the recommendation to change their diet and 711 patients who are prescribed meds are surveyed then there would be a 14.04% chance that the percent of the surveyed diet changers who lowered their cholesterol would be at least 2.6% greater than the percent of the surveyed med takers who lowered their cholesterol.
   • There is a 14.04% chance that patients who receive the change of diet recommendation are 2.6% more likely to lower their cholesterol than patients who are prescribed meds.
8. Interpret 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 681 patients who are given the recommendation to change their diet and 711 patients who are prescribed meds are surveyed then there would be a 5% chance that we would end up falsely concuding that the proportion of the 681 patients who received the diet change recommendation who lowered their cholesterol is greater than the proportion of the 711 patients who were prescribed meds who lowered their cholesterol.
   • There is a 5% 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 681 patients who are given the recommendation to change their diet and 711 patients who are prescribed meds are surveyed then there would be a 5% 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 greater than the success rate for the population of patients who are prescribed meds.

Answer 1

using excel

we have

Z Test for Differences in Two Proportions
Data
Hypothesized Difference	0
Level of Significance	0.05
Group 1
Number of Items of Interest	492
Sample Size	681
Group 2
Number of Items of Interest	495
Sample Size	711
Intermediate Calculations
Group 1 Proportion	0.72246696
Group 2 Proportion	0.696202532
Difference in Two Proportions	0.026264429
Average Proportion	0.7091
Z Test Statistic	1.0785
Upper Tail Test
Upper Critical Value	1.6500
p-Value	0.1404
Do not reject the null hypothesis

What can be concluded at the  = 0.05 level of significance?

1 )The null and alternative hypotheses would be:     

Ho:p1 = p2

Ha:p1 ≠ p2  

The test statistic z  = 1.079

The p-value = 0.1404

The p-value is >0.05

Based on this, we should fail to reject the null hypothesis.

Thus, the final conclusion is that ...

• The results are statistically insignificant at αα = 0.05, 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.

• There is a 14.04% chance of a Type I error.