In a survey, 45% of the respondents stated that they talk to their pets on the telephone. A veterinarian believed this result to be too high, so she randomly selected 200 pet owners and discovered that 84 of them spoke to their pet on the telephone. Does the veterinarian have a right to be skeptical? Use the a= 0.05 level of significance. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). Because npo (1-Po) = 10, the sample size is 5% of the population size, and the sample the requirements for testing the hypothesis satisfied. (Round to one decimal place as needed.) What are the null and alternative hypotheses? Ho (Type integers or decimals. Do not round.) versus H, Determine the test statistic, zo z0 = (Round to two decimal places as needed.) Determine the critical value(s). Select the correct choice below and fill in the answer box to complete your choice. (Round to two decimal places as needed.) O A. ±Za/2 = ± O B. Za = Does the veterinarian have a right to be skeptical? O A. The veterinarian does not have a right to be skeptical. There is not sufficient evidence to conclude that the true proportion of pet owners who talk to their pets on the telephone is less than 45%. O B. The veterinarian does not have a right to be skeptical. There is not sufficient evidence to conclude that the true proportion of t owners who talk to their on the telephone is 45%. O C. The veterinarian has a right to be skeptical. There is sufficient evidence to conclude that the true proportion of pet owners who talk to their pets on the telephone is less than 45%. O D. The veterinarian has a right to be skeptical. There is sufficient evidence to conclude that the true proportion of pet owners who talk to their pets on the telephone is not 45%.

In a survey of 1003 adults, a polling agency asked, "When you retire, do you think you will have enough money to live comfortably or not. Of the 1003 surveyed, 530 stated that they were worried about having enough money to live comfortably in retirement. Construct a 95% confidence interval for the proportion of adults who are worried about having enough money to live comfortably in retirement. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). Click here to view the table of critical t-values. Click here to view the table of critical values of the chi-square distribution. Select the correct choice below and fill in the answer boxes to complete your choice. (Use ascending order. Round to three decimal places as needed.) O A. There is a 95% probability that the true proportion of worried adults. between and O B. There is 95% confidence that the true proportion of worried adults is between and. OC. 95% of…

According to a study, the proportion of people who are satisfied with the way things are going in their lives is 0.86. Suppose that a random sample of 100 people is obtained. Complete parts (a) through (e) below. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). O A. The response is qualitative because the number of people satisfied can be counted. O B. The response is quantitative because the number of people satisfied can be counted. O C. The response is quantitative because the responses can be classified based on the characteristic of being satisfied or not. YD. The response is qualitative because the responses can be classified based on the characteristic of being satisfied or not. (b) Explain why the sample proportion, p, is a random variable. What is the source of the variability? The sample proportion p is a random variable because the value of p represents a random person included in the…

High school students across the nation compete in a financial capability challenge each year by taking a National Financial Capability Challenge exam, students who score in the top 19 percent are recognized publicly for their achievement by the Department of Treasury.  Assuming a normal distribution, how many standard deviations above the mean does a student have to score to be recognized?