Suppose a new treatment for a certain disease is given to a sample of 190 patients....

Suppose a new treatment for a certain disease is given to a sample of 190 patients. The treatment was successful for 151 of the patients. Assume that these patients are representative of the population of individuals who have this disease. Calculate a 98% confidence interval for the proportion successfully treated. (Round the answers to three decimal places.) Question: ____to____

Solutions

Expert Solution

Solution :

Given that,

Point estimate = sample proportion = = x / n = 151 / 190 = 0.795

1 - = 1 - 0.795 = 0.205

Z/2 = 2.326

Margin of error = E = Z / 2 * $\sqrt$ (( * (1 - )) / n)

= 2.326 * ( $\sqrt$ ((0.795 * 0.205) / 190)

Margin of error = E = 0.068

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.795 - 0.068 < p < 0.795 + 0.068

0.727 < p < 0.863

(0.727 to 0.863)

orchestra answered 1 year ago

Next > < Previous

Related Solutions

A certain medicine was given to a sample of 1800 patients. Of those patients given the...

A certain medicine was given to a sample of 1800 patients. Of those patients given the medicine, 845 had a significant improvement in their health status. Use that information to: a) Determine a 98% confidence interval for p, the population proportion of patients that would be helped by the medicine. b) Determine a 95% confidence interval for p, the population proportion of patients that would be helped by the medicine.

A study was done on 40 female patients following a new treatment for cardio-vascular disease (CVD). ...

A study was done on 40 female patients following a new treatment for cardio-vascular disease (CVD). Doctors measured the increase in exercise capacity (in minutes) over a 6-week period. The conventional treatment had produced an average increase of 2 minutes. Researchers wish to claim that the new treatment will increase the mean exercise capacity more than the conventional treatment. The data yielded x=2.3 and S =1.2 Conduct a 5-step test of hypotheses to test the researchers’ claim, use a =...

A small biotech company develops a new treatment for a rare disease. The new treatment is...

A small biotech company develops a new treatment for a rare disease. The new treatment is patented and the company is the sole monopolist in its market. The company can sell the treatment to private pharmacies and public hospitals. Pharmacies’ demand for the treatment is QPD = 84 – 0.4PP while public hospitals’ demand for the treatment is QHD = 116 – 0.6PH. The marginal cost of the new treatment is MC = 20 +2Q. The legislature passes a new...

Hemodialysis (also known as “dialysis”) is a common treatment for patients with kidney disease. In the...

Hemodialysis (also known as “dialysis”) is a common treatment for patients with kidney disease. In the United States, more than 468,000 patients are on dialysis (NIH, 2018). Since 1993, hospitalization rates related to the process of dialysis has risen between 47-87% (CDC, 2018). Many patients seek dialysis care at outpatient facilities that are not inspected as frequently as other healthcare facilities, like nursing homes and hospitals. For example, in California, kidney dialysis clinics are only inspected on average once every...

A study done at the Center for Disease at a certain university tracked 18,169 asymptomatic patients...

A study done at the Center for Disease at a certain university tracked 18,169 asymptomatic patients with a certain disease who started therapy at different points in the progression of the infection. It was discovered that asymptomatic patients who postponed antiretroviral treatment until their disease was more advanced faced a higher risk of dying than those who had initiated drug treatment earlier. Consider an experiment designed to determine the effectiveness of a new drug. The drug is given to participants...

A doctor administers an antibiotic to each of 250 patients. Of these patients, 190 recover without...

A doctor administers an antibiotic to each of 250 patients. Of these patients, 190 recover without need for further treatment. a. Find a point estimate for the proportion of all patients who would recover using this antibiotic. b. Find a 95% confidence interval for the proportion of all patients who would recover from this antibiotic. c. Suppose we wanted to repeat this study. How large a sample would we need if we wanted a 99% confidence interval with an error...

Age at diagnosis for each of 20 patients under treatment for meningitis was given in a...

Age at diagnosis for each of 20 patients under treatment for meningitis was given in a research paper. Suppose the ages (in years) were as follows. 18 18 27 19 23 20 66 18 21 18 20 18 18 20 18 19 28 16 18 18 (a) Calculate the values of the sample mean and the standard deviation. (Round your standard deviation to three decimal places.) sample mean = rstandard deviation = (b) Compute the upper quartile, the lower quartile,...

It is claimed that a new treatment of a very serious disease will substantially reduce the...

It is claimed that a new treatment of a very serious disease will substantially reduce the mortality of that disease from its present rate of 25%.We observe an SRS of 100 patients who have submitted to the new treatment and note that only 10 of them die from this disease. Does the new treatment seem to make a significant improvement?

A scientist designed a medical test for a certain disease. Among 100 patients who have the...

A scientist designed a medical test for a certain disease. Among 100 patients who have the disease, the test will show the presence of the disease in 96 cases out of 100, and will fail to show the presence of the disease in the remaining 4 cases out of 100. Among those who do not have the disease, the test will erroneously show the presence of the disease in 3 cases out of 100, and will show there is no...

Suppose that a the normal rate of infection for a certain disease in cattle is 25%....

Suppose that a the normal rate of infection for a certain disease in cattle is 25%. To test a new serum which may prevent infection, three experiments are carried out. The test for infection is not always valid for some particular cattle, so the experimental results are “inconclusive” to some degree–we cannot always tell whether a cow is infected or not. The results of the three experiments are (a) 10 animals are injected; all 10 remain free from infection. (b)...

Subjects