I am struggling to figure out this problem, not so much just for the answer but how exactly to do it. A researcher is interested in constructing a 99% confidence interval for the proportion of Alzheimerpatients living in nursing homes who exhibit temporary memory improvement while being visited bytheir loved ones. 920 of the 1356 randomly observed Alzheimer patients did show temporarymemory improvement.Round your answers to two decimal places.A. With 99% confidence the proportion of all Alzheimer patients living in nursing homes who exhibittemporary memory improvement while being visited by their loved ones is betweenandB. If many groups of 1356 randomly selected Alzheimer patients living in nursing homes areobserved while being visited by their loved ones, then a different confidence interval would beproduced for each group. Aboutpercent of these confidence intervals willcontain the true population proportion of all Alzheimer patients living in nursing homes that exhibittemporary memory improvement while being visited by their loved ones and aboutpercent will not contain the true population proportion.

Name: I am struggling to figure out this problem, not so much just for the a
Uploaded: 2024-03-20T06:38:44.000Z
Channel: Bartleby
Description: I am struggling to figure out this problem, not so much just for the answer but how exactly to do it. A researcher is interested in constructing a 99% confidence interval for the proportion of Alzheimerpatients living in nursing homes who exhibit tem

Given Information : A researcher is interested in constructing a 99% confidence interval for the proportion of Alzheimer patients living in nursing homes who exhibit temporary memory improvement while being visited by their loved ones. 920 of the 1356 randomly observed Alzheimer patients did show temporary memory improvement. We need to construct the 99% confidence interval for the population proportion. We have been provided with the following information about the number of favorable cases: Favorable Cases X = 920920 Sample Size N = 13561356 The sample proportion is computed as follows, based on the sample size N=1356 and the number of favorable cases X=920: The critical value for α=0.01 is zc=z1−α/2=2.576. The corresponding confidence interval is computed as shown below: Interpretation : Therefore, based on the data provided, the 99% confidence interval for the population proportion is 0.646<p<0.711, which indicates that we are 99% confident that the true population proportion pp is contained by the interval (0.646, 0.711). If many groups of 1356 randomly selected Alzheimer patients living in nursing homes are observed while being visited by their loved ones, then a different confidence interval would be produced for each group. About 99 percent of these confidence intervals will contain the true population proportion of all Alzheimer patients living in nursing homes that exhibit temporary memory improvement while being visited by their loved ones and about 1 percent will not contain the true population proportion.