i) Calculate an approximate 98% confidence interval with a conclusion for the true proportion of students that have problems with their health with continuous online classes.
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- A new television series must prove that it has more than 25% of the viewing audience after its initial 13-week run in order to be judged successful. Assume that in a sample of 400 households, 112 were watching the series. (a) Find the value of Za/2 in order to find a 99% confidence interval for the proportion of the viewing audience after its initial 13-week run.A website is trying to increase registration for first-time visitors, exposing 1% of these visitors to a new site design. Of 752 randomly sampled visitors over a month who saw the new design, 64 registered. Check any conditions required for constructing a confidence interval. Compute the standard error. Construct and interpret a 90% confidence interval for the fraction of first-time visitors of the site who would register under the new design (assuming stable behaviors by new visitors over time).A recent study surveyed adults on their views about the Covid-19 vaccine. In a random sample of 850 adults, 430 said that they are concerned that Covid-19 variants will hamper the country's ability to stop the spread of the disease. Use this sample to construct a 90% confidence interval for the population proportion of all U.S. adults who are concerned that Covid-19 variants will hamper the country's ability to stop the spread of the disease.
- The campus wellness center wants to estimate the proportion of students that are getting at least 7 hours of sleep every night during finals week, and they require an estimate with 99% confidence and an error margin of no more than 0.05. (a) If a similar survey from last year showed that p̂ = 0.38, what is the minimal number of students that must be surveyed this year? (b) If the center decides their data from last year is not comparable to this year due to the circumstances of online final exams, and so has no prior information regarding p̂, what is the minimal sample size needed?Suppose that in a simple random sample of 30 chicken carcasses from the ACME chicken plant (which has 10,000 chickens), 5 test positive for salmonella and 25 test negative for salmonella. Provide a point estimate and 95% CIs for the prevalence of salmonella among chicken carcasses from the ACME plant, assuming the test is 100% accurate. For the confidence interval, use a two-sided interval and report results from both the “exact method” and the “normal-theory method” (large sample approximation), then comment on whether the normal-theory method is reasonable based on comparison with results from the exact method.The trade magazine QSR routinely checks the drive-through service times of fast-food restaurants. An 80% confidence interval that results from examining 735 customers in Taco Bell's drive-through has a lower bound of 153.2 seconds and an upper bound of 158.2 seconds.
- A consumer watchdog group estimates the mean weight of 1-ounce fun size candy bars to see if customers are getting full value for their money. A random sample of 40 bars is selected and weighed, and the group reports that a 95% confidence interval for the true mean weight of the candy bars is 0.994 to 0.998 ounces. What is the margin of error?A trade magazine routinely checks the drive-through service times of fast-food restaurants. An 80% confidence interval that results from examining 591 customers in one fast-food chain's drive-through has a lower bound of 171.7 seconds and an upper bound of 175.1 seconds. What does this mean?A researcher is interested in testing whether, on average, College X graduates' earnings differ from those of College Y graduates who work in similar jobs and have similar characteristics. Using a matched pairs design that controls for all relevant characteristics, she finds that a 95% confidence interval for μCollegeX¯ μCollegeY is between -250 TL and 2500 TL . If the number of pairs chosen is 10, what can she conclude? A ) She can be 95% confident that there is a significant difference in the mean earnings between College X and College Y graduates. B ) She cannot be 99% confident that there is a significant difference in the mean earnings between College X and College Y graduates. C ) She can be 98% confident that there is a significant difference in the mean earnings between College X and College Y graduates. D ) She cannot be 80% confident that there is a significant difference in the mean earnings between College X and College Y graduates. E ) She cannot be 90% confident that…
- Suppose a researcher is interested in the effectiveness of a new drug intended to naturally increase hours of sleep for elderly patients experiencing insomnia. Suppose he collects information from a group of 64 patients and calculates a σ=.8 and a sample average of 6.4 hours of sleep. Is there sufficient evidence to reject the null hypothesis that his treatment is ineffective at increasing hours of sleep above 5 hours with 95% confidence?A consumer watchdog group estimates the mean weight of 1-ounce fun size candy bars to see if customers are getting full value for their money. A random sample of 40 bars is selected and weighed, and the group reports that a 95% confidence interval for the true mean weight of the candy bars is 0.994 to 0.998 ounces. What is the point estimate from this sample?1000 health practitioners in Zambia were surveyed during the COVID-19 outbreak, of these 380 of them though the Virus was not caused by people not wearing face masks. Find the margin or error for a 90% confidence interval with a critical value of 1.645