In a random sample of 373 people that were tested for the Norcovirus, it was found that 321 did not have the virus. A 94% confidence interval was constructed to estimate the proportion of the population that does not have the Norcovirus. This confidence interval came out to be (0.83,0.89).Select the correct intrepretation of this interval.We are 94% confident that the proportion of people in the sample that tested postitive for the Norcovirus is between 83% and 89%.We are 94% confident that the proportion of people in the sample that tested negative for the Norcovirus is between 83% and 89%.We are 94% confident that the proportion of the population that have the Norcovirus is between 83% and 89%.We are 94% confident that the proportion of the population that does not have the Norcovirus is between 83% and 89%.

As of October 2023 the current prevalence of Covid-19 in New South Wales is estimated to be roughly 1 in 1,000. Suppose a Rapid Antigen test has a 95% chance of correctly identifying someone with the virus and a 98% chance of correctly identifying someone without the virus. If we made all 8,000,000 people in New South Wales take a Rapid Antigen test what percentage of the people who tested positive would actually have Covid-19?Group of answer choices0.1%95%2.5%2%5.5%4.5%

A 95% confidence interval for the proportion of sea fans infected on the Las Redes Reef was calculated to be (0.423, 0.615)Based on this confidence interval, which one of the following statements is necessarily TRUE?Group of answer choices51.9% of all sea fans on the Las Redes Reef are infectedIf the study was repeated many times over, and the sample proportion of infected sea fans (p-hat) was recorded each time, then we would expect approximately 95% of the resulting estimates (the p_hat's) to be between (0.423 0.615).51.9% of all sea fans are infected.We don't know exactly what proportion of sea fans on the Las Redes Reef are infected but we know that it is definitely between 0.423 and 0.615.We can be 95% sure that the proportion of sea fans infected on the Las Redes Reef is between 0.423 and 0.615.

t is known that roughly 5% of the general population carry a certain virus. A health researcher believes that among caucasian males the percentage is actually lower. Which of the following will provide the most convincing evidence to support the researcher’s suspicion? Out of 25 randomly chosen caucasian males, 1 male carries the virus. Out of 70 randomly chosen caucasian males, 3 males carry the virus. Out of 6,000 randomly chosen caucasian males, 288 males carry the virus. Out of 6,000 randomly chosen caucasian males, 240 males carry the virus.

Researchers were interested in determining if grumpy old men were more likely to suffer heart disease than contented old men. A random sample of 300 men, aged between 55 and 65 years of age, was surveyed and their responses were recorded. Of the 100 grumpy old men surveyed 5 suffered from heart disease; of the 200 contented old men surveyed, 20 suffered from heart disease. Based on this information, which one of the following statements is FALSE?Group of answer choicesThe estimated difference in the incidence of heart disease for the two groups is -0.05.For constructing an approximate 95% confidence interval for (pgrumpy - pcontent ),the standard error is 0.0304.To test the hypothesis:   Ho : pgrumpy - pcontent = 0    against a two-sided alternative, the standard error will necessarily be greater than the standard error used for constructing the corresponding confidence interval.If we doubled the sample size, but still observed that 5% of the old grumpy men suffered from heart disease and 10% of the contented old men suffered from heart disease, then we would conclude that, for this age group, grumpy old men are (significantly) less likely to suffer from heat disease than contented men of a similar age.From this sample we can conclude that grumpy men, aged 55 to 65 years old, are (significantly) less likely to suffer from heart disease than contented men of a similar age.

Use the given data to construct a 99.5% confidence interval for the population proportion p.=x52,  =n71Round the answer to at least three decimal places.Theconfidenceintervalis, .