The medical researcher is comparing two treatments for lowering cholesterol: diet and meds. Theresearcher wants to see if the patients who receive the recommendation to change their diet have equalsuccess lowering cholesterol compared to a prescription of meds. A random sample of some patients whoreceived the recommendation to change their diet and others who were prescribed meds was taken. Theresults of how many did or did not lower their cholesterol are shown below:Data on Diet vs. Medsfor Weight LossDietYes 419No187What can be concluded at the α = 0.10 level of significance?For this study, we should use Select an answera. The null and alternative hypotheses would be:Ho: Select an answer ✓ Select an answer ✓H₁: Select an answerb. The test statistic ? | =c. The p-value:d. The p-value is ? ✓ ae. Based on this, we shouldf. Thus, the final conclusion is that …..=Select an answer ✓Meds446265Select an answerSelect an answer(please enter a decimal)(Please enter a decimal)(please show your answer to 3 decimal places.)(Please show your answer to 4 decimal places.)Select an answer the null hypothesis.O The results are statistically significant at a = 0.10, so there is sufficient evidence to concludethat the population of all patients who received the recommendation to change their diet isnot equally likely to lower their cholesterol as the population of patients who are prescribedmeds.O The results are statistically insignificant at a = 0.10, so we can conclude that the success ratefor all patients who receive the recommendation to change their diet is equal to the successrate for all patients who are prescribed meds.The results are statistically insignificant at a = 0.10, so there is insufficient evidence toconclude that the population of all patients who received the recommendation to change theirdiet is not equally likely to lower their cholesterol as the population of patients who areprescribed meds. g. Interpret the p-value in the context of the study.O If the success rate for the population of patients who receive the recommendation tochange their diet is the same as the success rate for the population of patients who areprescribed meds and if another 606 patients who are given the recommendation to changetheir diet and 711 patients who are prescribed meds are surveyed then there would be a1.46% chance that the percent of the surveyed diet changers who lowered their cholesterolwould differ at least 6.4% compared to the percent of the surveyed med takers wholowered their cholesterol.O If the success rate for the sample of patients who receive the recommendation to changetheir diet is the same as the success rate for the sample of patients who were prescribedmeds and if another 606 patients are given the recommendation to change their diet and711 patients are prescribed meds then there would be a 1.46% chance of concluding thatthe difference in the success rate for all patients who receive the recommendation tochange their diet and all patients who are prescribed meds is at least 6.4%.There is a 1.46% chance that the difference in the success rate for all patients who receivethe change of diet recommendation and all patients who are prescribed meds is at least6.4%.There is a 1.46% chance of a Type I error.h. Interpret the level of significance in the context of the study.There is a 10% chance that a patient won't be able to afford the meds, so they might aswell change their diet.O If the success rate for the population of patients who receive the recommendation tochange their diet is the same as the success rate for the population of patients who areprescribed meds and if another 606 patients who are given the recommendation to changetheir diet and 711 patients who are prescribed meds are surveyed then there would be a10% chance that we would end up falsely concuding that the success rate for thepopulation of patients who receive the recommendation to change their diet is differentfrom the success rate for the population of patients who are prescribed meds.O If the success rate for the population of patients who receive the recommendation tochange their diet is the same as the success rate for the population of patients who areprescribed meds and if another 606 patients who are given the recommendation to changetheir diet and 711 patients who are prescribed meds are surveyed then there would be10% chance that we would end up falsely concuding that the proportion of the 606 patientswho received the diet change recommendation who lowered their cholesterol is differentfrom the proportion of the 711 patients who were prescribed meds who lowered theircholesterol.

Given data: Diet Meds Total Yes 419 446 865 No 187 265 452 Total 606 711 1317 Significance level α=0.10 For this study we should use a z-test for the difference between two population proportions. (a) The null and alternative hypotheses would be: H0:p1=p2 H1:p1≠p2 (b) The test statistic is computed using the formula: z=p^1-p^2p^1-p^1n1+1n2 where, the sample proportions are: p^1=x1n1=419606=0.6914191419 p^2=x2n2=446711=0.6272855134 The pooled proportion is given by the formula: p^=x1+x2n1+n2=419+446606+711=0.6567957479 Substitute the values in the formula of the test statistic: z=p^1-p^2p^1-p^1n1+1n2=0.6914191419-0.62728551340.65679574791-0.65679574791606+1711=0.06413362850.2254150934×0.003056634777=2.443276065≈2.443 Thus, the value of the test statistic is 2.443 (c) The p-value can be obtained using Excel. Use the function "=2*(1-NORM.S.DIST(2.443,TRUE))". The result is given below: Thus, the required p-value is 0.0146 NOTE: Hi there! Thank you for the question. As per the honor code, we are allowed to answer three sub-parts at a time, so we are answering the first three as you have not mentioned which of these you are looking for. Please re-submit the question separately for the remaining sub-parts.