Find the critical value for a 95% confidence interval for B1 2. Compute the margin of error for 95% confidence interval 3. Contruct a 95% confidence interval

Echinacea is widely used as an herbal remedy for the common cold, but does it work? In a double-blind experiment, healthy volunteers agreed to be exposed to common-cold-causing rhinovirus type 39 and have their symptoms monitored. The volunteers were randomly assigned to take either a placebo or an echinacea supplement daily for 5 days following viral exposure. Among the 103 subjects taking a placebo, 88 developed a cold, whereas 44 of the 48 subjects taking echinacea 1 developed a cold.

Assume that a sample is used to estimate a population mean. Find the 90% confidence interval for a smaple of size 46 with a mean of 68.5 and a standard deviation of 6.6. Enter the answer as an open-interval (parentheses) accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

Find the critical value for a 98% confidence interval. Then calculate the interval.

Two different researchers measured the weight of two separate samples of ruby-throated hummingbirds from the same population. Each calculated a 95% confidence interval for the mean weight of these birds. Researcher 1 found the 95% confidence interval to be 3.12 g < μ < 3.48 g while Researcher 2 found the 95% confidence interval to be 3.05g < μ < 3.62 g.   Why would the two researchers get different answers? Which researcher most likely had the larger sample? Can you be certain about your answer in part (b)? Why or why not?

A researcher takes a random sample of 100 Americans that have never had a pet and a separate random sample of 100 Americans that are current or previous pet owners. She is interested in determining whether there is a difference in the proportion of individuals who believe in the afterlife among these two groups. She observes that 76 of the "never pet owners" and 84 of the "ever pet owners" believe in the afterlife. Determine the margin of error for a confidence interval for the difference in the two appropriate proportions at a 95% confidence level. a. The assumptions necessary to perform the calculation are not satisfied. b. 0.0837 c. 0.0719 d. 0.1103 e. 1.96

A consumer advocate analyses the nicotine content in some brand of cigarettes. A sample of 20 cigarettes of that Brand was analyzed which resulted in mean of 1.68mg. It is known that population standard deviation is 18. Nicotine content is assumed to be normally distributed. Find the 90% confidence interval for the true average nicotine content in that brand of cigarettes. (clearly write down the formula first and then calculate)Also, Interpret the meaning of confidence interval.

Suppose that for a random sample of 200 firms that revalued their fixed assets, the mean ratio of debt to tangible assets was 0.517 and the sample standard deviation was 0.148. For an independent random sample of 400 firms that did not revalue their fixed assets, the mean ratio of debt to tangible assets was 0.489 and the sample standard deviation was 0.158. Find a 99% confidence interval for the difference between the two population means.