An executive believes that no more than 80% of the company’s employees take all of their vacation days. In a sample of 210 members, 158 employees took all of their vacation days. Therefore, the p-value is 0.0422. When testing the executive's (using a 5% level of significance), what can you conclude concerning the null hypothesis? Reject the null hypothesis Fail to reject the null hyp
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- An executive believes that no more than 80% of the company’s employees take all of their vacation days. In a sample of 210 members, 158 employees took all of their vacation days. Therefore, the p-value is 0.0422. When testing the executive's (using a 5% level of significance), what can you conclude concerning the null hypothesis?
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Reject the null hypothesis |
||
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Fail to reject the null hypothesis |
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- In the past, 30% of a country club's members brought guests to play golf sometime during the year. Last year, the club initiated a new program designed to encourage members to bring more guests to play golf. In a sample of 95 members, 34 brought guests to play golf after the program was initiated. When testing the hypothesis that the new program has increased the proportion of members bringing out guests (using a 5% level of significance), what is the null and alternative hypothesis?A marriage counselor has traditionally seen that the proportion p of all married couples for whom her communication program can prevent divorce is 77%. After making some recent changes, the marriage counselor now claims that her program can prevent divorce in more than 77% of married couples. In a random sample of 250 married couples who completed her program, 194 of them stayed together. Based on this sample, is there enough evidence to support the marriage counselor's claim at the 0.10 level of significance? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.) (a) State the null hypothesis H and the alternative hypothesis H₁. H₁:0 H₁ :0 (b) Determine the type of test statistic to use. (Choose one) (c) Find the value of the test statistic. (Round to three or more decimal places.) 0 (d) Find the p-value. (Round to three or more decimal places.) 0 (e) Is there enough…A newsletter publisher believes that less than 36%36% of their readers own a laptop. For marketing purposes, a potential advertiser wants to confirm this claim. After performing a test at the 0.010.01 level of significance, the advertiser failed to reject the null hypothesis. What is the conclusion regarding the publisher's claim?
- US Universities found that 72% of people are concerned about the possibility that their personal records could be stolen over the internet. If a random sample of 300 college students at a Midwestern university were taken and 228 of them were concerned about the possibility that their personal records could be stolen over the Internet, could you conclude at the 0.025 level of significance that a higher proportion of the university’s college students are concerned about Internet theft than the public at large? Report the p-value for this test. Z0.025 = 1.96A marriage counselor has traditionally seen that the proportion p of all married couples for whom her communication program can prevent divorce is 79%. After making some recent changes, the marriage counselor now claims that her program can prevent divorce in more than 79% of married couples. In a random sample of 240 married couples who completed her program, 194 of them stayed together. Based on this sample, is there enough evidence to support the marriage counselor's claim at the 0.10 level of significance? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.) (a) State the null hypothesis Ho and the alternative hypothesis H. H. :0 = 0.79 H, : o > 0.79 (b) Determine the type of test statistic to use. OSO (c) Find the value of the test statistic. (Round to three or more decimal places.) OUSA Today reported that about 47% of the general consumer population in the United States is loyal to the automobile manufacturer of their choice. Suppose Chevrolet did a study of a random sample of 1007 Chevrolet owners and found that 482 said they would buy another Chevrolet. Does this indicate that the population proportion of consumers loyal Chevrolet is more than 47%? Use a = 0.01. (a) What is the level of significance? State the null and alternate hypotheses. O Ho: P = 0.47; H,: p 0.47 O Ho: p > 0.47; H,: p = 0.47 O Ho: p = 0.47; H1: p + 0.47 (b) What sampling distribution will you use? O The Student's t, since np 5 and ng > 5. O The standard normal, since np 5 and ng > 5. What is the value of the sample test statistic? (Round your answer to two decimal places.) (c) Find the P-value of the test statistic. (Round your answer to four decimal places.) Sketch the sampling distribution and show the area corresponding to the P-value.In a survey, 37% 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 200 pet owners and discovered that 73 of them spoke to their pet on the telephone. Does the veterinarian have a right to be skeptical? Use the a = 0.1 level of significance. 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.) Enter your answer in the answer box and then click Check Answer. ? 4 parts remaining Clear All Check Answer Screen Shot 202...png Show AllA marriage counselor has traditionally seen that the proportion p of all married couples for whom her communication program can prevent divorce is 78%. After making some recent changes, the marriage counselor now claims that her program can prevent divorce in more than 78% of married couples. In a random sample of 240 married couples who completed her program, 189 of them stayed together. Based on this sample, is there enough evidence to support the marriage counselor’s claim at the 0.10 level of significance?Perform a one-tailed test. Then complete the parts below.One study claimed that 88 % of college students identify themselves as procrastinators. A professor believes that the claim regarding college students is too high. The professor conducts a simple random sample of 272 college students and finds that 231 of them identify themselves as procrastinators. Does this evidence support the professor's claim that fewer than 88 % of college students are procrastinators? Use a 0.02 level of significance. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. Ho p = 0.88 Ha P 0.88A nationwide survey of working adults indicates that out of 100 adults, 50 of them are satisfied with their jobs. The president of a large company believes that more than this number of employees at his company are satisfies with their jobs. To test his belief, he surveys a random sample of 100 employees, and 59 of them report that they are satisfied with their jobs. Do you support president's statement about the employee’s satisfaction? (Use a= 0.05 level of significance.) Find H0 and Ha. Find Test Statistic Identify Decision Identify ConclusionOne study claimed that 85 % of college students identify themselves as procrastinators. A professor believes that the claim regarding college students is too high. The professor conducts a simple random sample of 108 college students and finds that 86 of them identify themselves as procrastinators. Does this evidence support the professor's claim that fewer than 85 % of college students are procrastinators? Use a 0.01 level of significance. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. Ho :p = 0.85 0.85 d: "H Answer 囲 Tables 國 Keypad Keyboard Shortcuts O > bAdults randomly selected for a poll were asked if they "favor or oppose using federal tax dollars to fund medical research using stem cells obtained from human embryos." Of the subjects surveyed, 257 were in favor, 281 were opposed, and 103 were unsure. A politician claims that people don't really understand the stem cell issue and their responses to such questions are random responses equivalent to a coin flip. Use a 0.01 significance level to test the claim that the proportion of subjects who respond in favor is equal to 0.5. Find the null and alternative hypothesis. H0: p=0.5 p<0.5 p≠0.5 p>0.5 H1: p<0.5 p>0.5 p≠0.5 p=0.5 If we consider + to represent those in favor, then how many of each sign is there? Positive Signs: Negative Signs: Total Signs: What is the p-value? (Round to three decimal places.) What is the conclusion about the null? What is the conclusion about the claim?Recommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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