1. (06.01 LC) An opinion poll asked a random sample of high school couples in the United States whether they believe they will marry their high school sweetheart. A researcher believes more than 25% of high school couples believe they will marry. Which null and alternative hypotheses should be used to test this claim? (4 points) O Ho: P = 0.25, Ha: P > 0.25 Ho: P = 0.25, Ha: P 0.25, Ha: P = 0.25 O Ho: p = 0.25, Ha: P > 0.25 2. (06.01 MC) A student performs a test of Ho: p = 0.6 versus Ha: p > 0.6 and gets a p-value of 0.75. The student writes: "Because the p-value is greater than 0.6, we reject the null hypothesis." What is wrong with the student's conclusion? (4 points) The p-value should be compared with the mean, not the hypothesized value of p. The p-value should be compared with the standard deviation, not the hypothesized value of p. O The p-value should be compared with a significance level such as a = 0.05, not the hypothesized value of p. O The student should have stated the data prove Ha is true. O The student should have stated the data prove Ho is true.11. (06.05 LC) You want to find whether there is a difference in the proportion of male and female college students who had a job during the past semester. Out of a simple random sample of 200 female and 250 male college students, you find that 146 females and 212 males reported they had a job the previous semester. Which hypotheses should you test if Pm and pf denote the proportions of all college males and females, respectively, who worked the previous semester? (5 points) Ho: Pm - Pf = 0 versus Ha: Pm - Pf = 0 Ho: Pm - Pf = 0 versus Ha: Pm - Pf > 0 O Ho: Pm - Pf = 0 versus Ha: Pm - Pf 0 versus Ha: Pm - Pf = 0 Ho: Pm - Pf = 0 versus Hai Pm - Pf = 012. (06.05 MC) Which of the following is a TRUE statement about hypothesis testing? (4 points) If there is sufficient evidence to reject a null hypothesis at the 10% level, then there is sufficient evidence to reject it at the 5% level. Whether to use a one-sided or a two-sided test is typically decided before the data are gathered. If a hypothesis test is conducted at the 1% level, there is a 1% chance of rejecting the null hypothesis. The probability of a Type I error plus the probability of a Type II error always equals one. O The power of a test concerns its ability to detect a null hypothesis.3. (06.02 LC) In a test of the hypothesis H0: 0 : 0.8 against Ha: p > 0.8, the power of the test when p 2 0.9 is greatest for which of the following? (4 points) @ n: 75.0: 0.05 O n: 75.0: 0.10 O H: 70.11: 0.10 O H: 70.11: 0.05 0 There is not enough information given to draw a conclusion. 4. (06.02 MC) Suppose a political advisor is interested in the proportion of the vote an opponent will receive. If he samples voters randomly and tests hypotheses regarding p, the population proportion, what should he do to reduce his risk of making a Type II error? (4 points) I. Increase the number of voters he will sample Il. Decrease the number of voters he will sample Ill. Increase the significance level IV. Decrease the significance level I only O ll only O Ill only II and IV only I and Ill only5. (06.02 MC) Members of the homeowners' association (HOA) want to know whether more than 45% of homeowners from a local residential community supports a 2% annual increase in HOA fees for five years to fund the resurfacing of the community swimming pool. The HOA surveys a random sample of 50 homeowners and use the results to test the hypotheses Ho: p = 0.45 and Ha: p > 0.45, where p is the proportion of all homeowners who support a 2% annual increase in HOA fees for five years to fund the pool resurfacing. In the context of this study, how could a Type I error occur? (4 points) The HOA finds evidence that more than 45% of homeowners do not support the fee increase, when there isn't convincing evidence that more than 45% supports the increase. O The HOA finds evidence that more than 45% of homeowners support the fee increase, when at most, 45% of the homeowners support the increase. O The HOA finds evidence that more than 45% of homeowners support the fee increase, when more than 45% of the homeowners do support the increase. O The HOA does not find evidence that more than 45% of homeowners support the fee increase, when more than 45% of the homeowners do support the increase. The HOA does not find evidence that more than 45% of homeowners support the fee increase, when at most, 45% of the homeowners do support the increase. 6. (06.02 MC) An individual is arrested for stealing and pleads not guilty at trial. The jury can only find him guilty if there is evidence beyond a reasonable doubt to reject the assumption of innocence. Describe what could occur if the jury makes a Type | error? (4 points) The jury finds the individual innocent and he is released from jail, but he is actually guilty. The jury finds the individual guilty, he is put in jail, and he is actually guilty. O The jury finds the individual innocent, he is released from jail, and he is actually innocent. The jury finds the individual guilty and he is put in jail, but he is actually innocent. A conclusion cannot be made based on the information given.7. (06.04 MC) A poll of 1,000 randomly selected registered voters was taken and 584 responded that they favor candidate X for governor (p 1 = 0.5840). Just before the election, another poll of 950 registered voters was taken and 401 individuals responded that they favor candidate X (P 2 = 0.4221). A 95% two-proportion z confidence interval for the true difference between p , and p 2 was found to be (0.1181, 0.2057). What is the meaning of the interval in the context of the problem? (4 points) There has been a substantial gain in support for the candidate. O There has been a substantial drop in support for the candidate. O There has been no change in support for the candidate. O It is probable for the candidate to get most of the votes because the confidence interval is positive. It is not probable for the candidate to get most of the votes because support has dropped below 50%.8. (06.04 LC) A recent study states that moonlight can affect the survival rate of new plants. It reported that 26 of 50 plants that were shielded from moonlight survived and 19 of 65 plants left unshielded survived. Choose the 95% confidence interval that represents the difference between the proportions of shielded and unshielded new plants that survive. (4 points) 26 26 19 19 50 50 65 65 26 19 + 1.961 50 65 50 O 26 26 19 19 50 50 65 65 20 19 + 1.6451 50 65 50 19 45 65 + 1.96 50 65 115 115 50 65 65 26 19 45 + 1.645 50 65 115 115 50 65 O 26 19 45 65 26 19 + 1.96 50 65 115 115 50 65 9. (06.05 MC)9. (06.05 MC) A researcher wants to conduct a study to determine whether a newly developed anti-tardy program is successful. Two random groups of 100 students each, identified as control and treatment groups, are formed from 200 students who are repeatedly late to school. Both groups receive a set of anti-tardy reading materials and a lecture from a teacher and a tardy-reformed student about the negative impacts of being late. In addition, the treatment group also receives materials from the newly developed program. Thirty-three of the 100 students in the control and 35 of the 100 students in the treatment group are no longer late to school is recorded 6 months later. Use the results to test the hypotheses Ho: P1 = P2 and Ha: P1 = P2 where p, represents the proportion of students who succeed in the control program and p2 represents the proportion of students who succeed in the newly developed program. What can you conclude using the significance level a = 0.05? (5 points) The new program is not significantly different than the control program at reducing tardiness because the p-value is greater than a = 0.05. The new program is effective at reducing tardiness because the proportion of students who are no longer late is greater in the treatment group than control group. The new program is significantly different than the control program at reducing tardiness because the p-value is greater than a = 0.05. The new program is not significantly different than the control program at reducing tardiness because the difference in the proportions of the control group and treatment group is small. The new bullying program is effective at reducing bullying because the standard deviation is small, 0.0141.10. (06.05 MC) The way in which response options are presented in a question can affect a person's response. Two hundred randomly selected people were asked about their milk chocolate or dark chocolate preference. One hundred of the participants were randomly given the option of milk chocolate first and the remaining 100 participants were given the option of dark chocolate first. The results are given in the table. Milk Chocolate Dark Chocolate Milk chocolate option first 52 48 Dark chocolate option first 41 59 To conclude if the order in which options are presented in a question affects the answer, a two-proportion z-test was conducted. What is the correct p-value of the test? (4 points) 0.0594 1 0.1189 0.3193 0.4650 0.5200