Zarina wants to determine a confidence interval based on a random poll she conducted about the percent of attendees at a conference who are vegetarian. she determined the margin of error for her poll results using the equation below. e = 1.96 /.3(1-.3)/540 based on zarina’s work, what can be concluded? with 90% confidence, between 4% and 30% of the attendees are vegetarian. with 90% confidence, between 26% and 34% of the attendees are vegetarian. with 95% confidence, between 4% and 30% of the attendees are vegetarian. with 95% confidence, between 26% and 34% of the attendees are vegetarian.

Acell phone company offers two plans to its subscribers. at the time new subscribers sign up, they are asked to provide some demographic information. the mean yearly income for a sample of 40 subscribers to plan a is $45,000 with a standard deviation of $9,200. for a sample of 25 subscribers to plan b, the mean income is $64,300 with a standard deviation of $7,100. at the 0.025 significance level, is it reasonable to conclude the mean income of those selecting plan b is larger? assume unequal population standard deviations. hint: for the calculations, assume the plan a as the first sample. what is the decision rule? (negative amount should be indicated by a minus sign. round your answer to 3 decimal places.) compute the value of the test statistic. (negative amount should be indicated by a minus sign. round your answer to 2 decimal places.) what is your decision regarding h0 ? reject h0 fail to reject h0 compute the p-value. less than 0.01 between 0.01 and 0.025 between 0.025 and 0.05 between 0.05 and 0.10 greater than 0.10