A random sample of 360 married couples found that 280 had two or more personality preferences in common. In another random sample of 560 married couples, it was found that only 32 had no preferences in common. Let p1 be the population proportion of all married couples who have two or more personality preferences in common. Let p2 be the population proportion of all married couples who have no personality preferences in common.   (a) Find a 95% confidence interval for p1 – p2. (Use 3 decimal places.) lower limit       upper limit      (b) Explain the meaning of the confidence interval in part (a) in the context of this problem. Does the confidence interval contain all positive, all negative, or both positive and negative numbers? What does this tell you (at the 95% confidence level) about the proportion of married couples with two or more personality preferences in common compared with the proportion of married couples sharing no personality preferences in common?   a. Because the interval contains only positive numbers, we can say that a higher proportion of married couples have two or more personality preferences in common.   b. Because the interval contains both positive and negative numbers, we can not say that a higher proportion of married couples have two or more personality preferences in common.       c. Because the interval contains only negative numbers, we can say that a higher proportion of married couples have no personality preferences in common.   d. We can not make any conclusions using this confidence interval.

When Mendel conducted his famous genetics experiments with peas, one sample of offspring consisted of 428 green peas and 152 yellow peas. a.Find a 99% confidence interval estimate of the percentage of yellow peas b. Based on his theory of genetics, Mandel expected that 25% of the offspring peas would be yellow. Given that percentage of offspring yellow peas is not 25%, do  the results contradict mendel's theory? Why or why not?

A newsgroup is interested in constructing a 95% confidence interval for the difference in the proportions of Texans and New Yorkers who favor a new Green initiative. Of the 528 randomly selected Texans surveyed, 432 were in favor of the initiative and of the 554 randomly selected New Yorkers surveyed, 455 were in favor of the initiative. a. With 95% confidence the difference in the proportions of Texans and New Yorkers who favor a new Green initiative is between (round to 3 decimal places) and (round to 3 decimal places). b. If many groups of 528 randomly selected Texans and 554 randomly selected New Yorkers were surveyed, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population proportion of the difference in the proportions of Texans and New Yorkers who favor a new Green initiative and about percent will not contain the true population difference in proportions. Ouestion Help: M Message…

A survey was conducted to determine what percentage of college seniors would have chosen to attend a differentcollege if they had known then what they know now. In a random sample of 100 seniors, 34 percent indicated thatthey would have attended a different college. A 90 percent confidence interval for the percentage of all seniors whowould have attended a different college is

A random sample of Roosevelt High School students was asked if they prefer taking notes on a computer or using a paper and pencil. Of the 180 students who were surveyed, 75 said they preferred using the computer. The resulting 99 percent confidence interval for the proportion of students who prefer taking notes on the computer was (0.322, 0.511). The school newspaper ran a story saying that less than half of the student body prefers taking notes on a computer. Based on the interval, is the newspaper justified in its statement? Yes. The sample proportion of students who prefer taking notes on their computer was 0.417, which is less than 0.5. Yes. Since nearly the entire confidence interval contains values less than 0.5, we can be certain that less than 50% of students prefer using a computer for notes. No. Since the interval contains values above 0.5, it is plausible that more than 50% of students prefer using a computer for notes. No. Since the interval contains values above 0.5, we…

You are interested in constructing a 90% confidence interval for the proportion of all caterpillars that eventually become butterflies. Of the 427 randomly selected caterpillars observed, 45 lived to become butterflies. a.  Use 3 decimal places to complete the following statement: With 90% confidence the proportion of all caterpillars that lived to become a butterfly is between  and . b.  If many groups of 427 randomly selected caterpillars were observed, then a different confidence interval would be produced from each group. About  percent of these confidence intervals will contain the true population proportion of caterpillars that become butterflies and about  percent will not contain the true population proportion.

In a survey funded by the UW school of medicine, 750 of 1000 adult Seattle residents said they did not believe they could come down with a sexually transmitted infection (STI). Construct a 95% confidence interval estimage of the proportion of adult Seattle residents who don't believe they can contract an STI. (Use a z score of 1.96 for your computations.) (.728, .772) (.723, .777) (.718, .782) (.713, .878) (.665, .835)

5.A random sample of 362 married couples found that 286 had two or more personality preferences in common. In another random sample of 550 married couples, it was found that only 30 had no preferences in common. Let p1 be the population proportion of all married couples who have two or more personality preferences in common. Let p2 be the population proportion of all married couples who have no personality preferences in common. Find a 90% confidence interval for p1 – p2. (Use 3 decimal places.) lower limit       upper limit

In September 2011, Gallup surveyed 1,004 American adults and asked them whether they blamed Barack Obama a great deal, a moderate amount, not much, or not at all for U.S. economic problems. The results showed that 53% of respondents blamed Barack Obama a great deal or a moderate amount. Calculate the 99% confidence interval for the proportion of all American adults who blame Barack Obama a great deal or a moderate amount for U.S. economic problems. O (0.504, 0.556) O (0.499, 0.561) O (0.489, 0.571) O The confidence interval cannot be determined from the information given.