In a survey, 36% of the respondents stated that they talk to their pets on the telephone. A veterinarian believed this result to be too high, so he randomly selected 240 pet owners and discovered that 81 of them spoke to their pet on the telephone. Does the veterinarian have a right to be skeptical? Use the a = 0.01 level of significance. Because npo (1- Po) = 10, the sample size is v 5% of the population size, and the sample the V satisfied. requirements for testing the hypothesis (Round to one decimal place as needed.)

Consumer Reports did a survey of 1100 adults and found that 565 were unhappy that their food did not look like the picture in the menu. Use a 0.05 significance level to test the claim that the majority of adults are annoyed by food that does not look like the picture in the menu. What is the test statistic for this problem? TTTArial 3 (12pt) v T- E

The US Department of Energy reported that 48% of homes were heated by natural gas. A random sample of 331 homes in Oregon found that 132 were heated by natural gas. Test the claim that proportion of homes in Oregon that were heated by natural gas is different than what was reported. Use a 1% significance level. Give answer to at least 4 decimal places. What are the correct hypotheses? (Select the correct symbols and use decimal values not percentages.) H0:                   H1:                     Based on the hypotheses, compute the following:Test Statistic = p-value =  Based on the above we choose to       The correct summary would be:       that the proportion of homes in Oregon that were heated by natural gas is different than what the DOE reported value of 48%.

Suppose an instructor would like to know if there is difference in the performance of her two courses of 560 students each. She proceeds to take a random sample of size 50 from each independent class and was able to establish the following results: mean SD Class 1 86 17 Class 2 78 15 Is there a significant difference between the two classes at a significance level of five percent?

A shareholders' group is lodging a protest against your company. The shareholders group claimed that the mean tenure for a chief exective office (CEO) was at least 11 years. A survey of 62 companies reported in The Wall Street Journal found a sample mean tenure of 9.3 years for CEOS with a standard deviation of s = 4.4 years (The Wall Street Journal, January 2, 2007). You don't know the population standard deviation but can assume it is normally distributed. You want to formulate and test a hypothesis that can be used to challenge the validity of the claim made by the group, at a significance level of a = 0.05. Your hypotheses are: H.:µ > 11 На: :и < 11 What is the test statistic for this sample?

In the General Social Survey, respondents were asked "If your party nominated a woman for President, would you vote for her if she were qualified for the job?" A two-way table summarizing the results for 953 respondents, by gender, is shown below: Gender Female Male Total Vote for female candidate? Yes No 488 66 335 64 823 130 The chi-square test statistic = 3.35. At a significance level of 0.05, what is your conclusion? O The null hypothesis is not rejected: there is no association between gender and support for a female president Total 554 399 953 O The null hypothesis is rejected: there is an association between gender and support for a female president. O The null hypothesis is not rejected: there is an association between gender and support for a female president. ty O The null hypothesis is rejected: there is no association between gender and support for a female president. Q Search or enter address +

A dietitian wishes to see if a person's cholesterol level will change if the diet is supplemented by a certain mineral. Seven subjects were pretested and then took the mineral supplement for a six-week period. The results are shown below in the table. Use a paired samples t-chart at a = 0.01 significance level to see difference (Before - After) between the cholesterol levels. State the hypotheses and conclusion. Also, conduct a 99% confidence interval. Subject 1 Before 210 After 190 2 205 170 3 208 210 4 192 188 5 178 173 6 244 228 7 211 198

A researcher assures that in his country, on average, the salary of workers is less than $5000.00. He takes a sample of 7 workers and finds that their salary is: $3900.00, $5100.00, $3100.00, $5500.00, $2700.00, $4500.00, $4100.00 Can you be sure at the 0.02 level of significance that the researcher is correct? Choose 1. The correct conclusion2. The correct interpretation Select one:to)1. H0 is rejected2. With a confidence of 98% it can be ensured that the average wage of the workers is not less than $5000.00b)1. H0 is not rejected2. With a confidence of 98% it can be ensured that the average wage of the workers is not less than $5000.00c)1. H0 is not rejected2. It can be ensured, with 98% confidence, that the average wage of the workers is less than $5,000.00d)1. H0 is rejected2. It can be ensured, with 98% confidence, that the average wage of the workers is less than $5,000.00

Suppose that in a random selection of 100 colored​ candies, 23​% of them are blue. The candy company claims that the percentage of blue candies is equal to 26​%. Use a 0.05 significance level to test that claim.

Previously, an organization reported that teenagers spend 12.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is lower. Twenty-three randomly chosen teenagers were asked how many hours per week they spend on the phone, with the following results (in hours): 10.4, 10.4, 17.3, 11.2, 10, 15.2, 9.5, 5.8, 12.5, 11.8, 7.3, 10.6, 11.5, 16.2, 16.5, 14.3, 12.3, 9.9, 13.7, 8.6, 11.3, 13, 16.5 Perform a hypothesis test using a 10% level of significance. Step 1: State the null and alternative hypotheses. Ho: pv 12.5 12.5 (So we will be performing a left-tailed Vv test.) Step 2: Assuming the null hypothesis is true, determine the features of the distribution of point estimates using the Central Limit Theorem. By the Central Limit Theorem, we know that the point estimates are t-distributed with distribution mean and distribution standard deviation Step 3: Find the p-value of the point estimate. = P(tv sv p-value = Step 4: Make a Conclusion About the nulI…