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
Q: A researcher designs an experiment in which the single independent variable has five levels. If the…
A: Given A researcher designs an experiment in which the single independent variable has five levels.…
Q: A random sample of 860 births in New york state included 434 girls. Use a 0.05 significance level to…
A: The aim is to test the claim that 48.8% of newborn babies are girls. The test hypotheses are given…
Q: Is there enough evidence to support the college's claim at 5% level of significance?
A: It is given that Sample size n = 100 Number of unemployed, X = 27 Level of significance = 0.05
Q: A statistics student believes that black cars are less likely to receive tickets for moving…
A: It is given that the P-value is 0.24.
Q: in a Gallup poll of 1003 randomly selected subjects, 370 said they have a gun in their home. The…
A: The z statistic is Z=(p-mean)/standard deviation Thus, the z statistic is -2.01
Q: Suppose 450 subjects are treated with a drug that is used to treat pain and 140 of them developed…
A: The sample size, n is 450 and the number of patients who have developed nausea is 140. The sample…
Q: Small-business telephone users were surveyed 6 months after access to carriers other than Carrier A…
A:
Q: One research study of illegal drug use among 12- to 17-year-olds reported a decrease in use (from…
A: Givensample size(n)=1050x=95significance level(α)=0.05
Q: A marriage counselor has traditionally seen that the proportion p of all married couples for…
A:
Q: Historically, 15% of all Americans claim to have seen a UFO. A paranormal researcher claims that…
A: Consider that p is the true proportion of Americans, who claimed to have seen a UFO.
Q: A marriage counselor has traditionally seen that the proportion p of all married couples for whom…
A: From the provided information, Sample size (n) = 245 From which 189 of them stayed together. Sample…
Q: Adults randomly selected for a poll were asked if they "favor or oppose using federal tax dollars to…
A: GIVEN DATA, 451- in favor450- opposed102-unsure 1003- total peopleα=0.05 claim:p=0.50x= no. of in…
Q: A 2018 poll of 3619 randomly selected users of a social media site found that 2463 get most of their…
A: given data (a) claim : p ≠ 0.5n = 3619x = 2463α = 0.05p^ = xn = 24633619 = 0.6806
Q: Adults randomly selected for a poll were asked if they "favor or oppose using federal tax dollars to…
A:
Q: Small-business telephone users were surveyed 6 months after access to carriers other than AT&T…
A: The question is about hypo. testing Given : Random sample of users ( n1 ) = 368 No. of users who…
Q: In one study of smokers who tried to quit smoking with nicotine patch therapy, 38 were smoking one…
A: Total sample size, n = 38 + 34 = 72 Smokers who tried to quit smoking with nicotine patch therapy,…
Q: In one study of smokers who tried to quit smoking with nicotine patch therapy, 39 were smoking one…
A: The information provided in the question is as follows :-Number of people who were smoking after the…
Q: Market Research, Inc., wants to know if shoppers are sensitive to the prices of items sold in a…
A: The question is about hypo. testing Given : Randomly selected no. of shoppers ( n ) = 802 No. of…
Q: According to a geographic report, in the United States, 20% of all families renting condos will move…
A:
Q: A researcher thinks that watching violent televisions increases aggressive behavior in children. To…
A: Given 48 children 3 groups
Q: College-aged adults need at least 7 hours of sleep each night to stay healthy. Sleep deprivation can…
A:
Q: Twenty years ago, 51% of parents of children in high school felt it was a serious problem that high…
A: Givenn=850P0=0.51so n*P0(1-P0) = 850*0.51*(1-0.51) = 212.4
Q: A national publishing house claims that 35% of all weekly magazine readers inSouth Africa read their…
A: Null Hypothesis: H0: The proportion of all weekly magazine readers in South Africa read their…
Q: One study claimed that 86 % of college students identify themselves as procrastinators. A professor…
A: Answer Sample size [n] =139favorable cases [x] =110
Q: In a study of smokers who tried to quit smoking with nicotine patch therapy, 39 were smoking after…
A: Given Information: The number of smokers who tried to quit smoking with nicotine patch therapy,…
Q: We are interested in testing whether the proportion of American adults who mostly drink diet soda…
A: X=485, n=2027, level of significance ɑ=0.05 ,p=25%
Q: In a random sample of 315 people 90 of them were smokers. Based on the above sample test whether the…
A:
Q: n February 2008, an organization surveyed 1040 adults age 18 and older and found that 539 believe…
A: State the hypotheses.
Q: he national membership for the CPAs,
A: The sample size is N=620. The number of favorable cases is X=63. The sample proportion is , and the…
Q: Adults randomly selected for a poll were asked if they "favor or oppose using federal tax dollars…
A:
Q: Suppose that in a random selection of 100 colored candies, 29% of them are blue. The candy company…
A: The information provided in the question are as follows :- Sample size (n) = 100 The sample…
Q: USA Today reported that about 47% of the general consumer population in the United States is loyal…
A: We have given that the hypothetical proportion p0=0.47, sample size 1002 and favourable cases X =…
Q: It is believed that 11% of all Americans are left-handed. A college needs to know how many…
A: Given that Sample size n =200 Favorable cases x =30 Sample proportion p^=x/n =30/200 =0.15
Q: Adults randomly selected for a poll were asked if they "favor or oppose using federal tax dollars…
A:
Q: A cancer research group surveys a random sample of 400 women more than forty years old to test the…
A: Sample size , n = 400 women more than forty years old 140 women in the sample respond affirmatively…
Q: (a) State the null hypothesis Ho and the alternative hypothesis H₁. H:0 H₁:0 (b) Determine the type…
A: It is given that Population proportion, p = 80% = 0.80 Favourable cases, X = 178 Sample size, n =…
Q: For a certain year, a study reports that the percentage of college students using credit cards was…
A:
Q: With the null hypothesis that a suspect is innocent, an investigation concluded that the suspect…
A: The probability of determining a test statistic value at least as extreme as the observed value…
- 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 hypothesis |
Step by step
Solved in 2 steps with 2 images
- 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 > bA large manufacturing company claims that less than 10% of its customers make a complaint about the company’s products. Test this claim at the 10% significance level, if a survey found that 29 out of a random sample of 320 of the company’s customers made a complaint about the company’s products.SEE MORE QUESTIONSRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th…StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage LearningStatistics for The Behavioral Sciences (MindTap C…StatisticsISBN:9781305504912Author:Frederick J Gravetter, Larry B. WallnauPublisher:Cengage LearningElementary Statistics: Picturing the World (7th E…StatisticsISBN:9780134683416Author:Ron Larson, Betsy FarberPublisher:PEARSONThe Basic Practice of StatisticsStatisticsISBN:9781319042578Author:David S. Moore, William I. Notz, Michael A. FlignerPublisher:W. H. FreemanIntroduction to the Practice of StatisticsStatisticsISBN:9781319013387Author:David S. Moore, George P. McCabe, Bruce A. CraigPublisher:W. H. FreemanMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th…StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage LearningStatistics for The Behavioral Sciences (MindTap C…StatisticsISBN:9781305504912Author:Frederick J Gravetter, Larry B. WallnauPublisher:Cengage LearningElementary Statistics: Picturing the World (7th E…StatisticsISBN:9780134683416Author:Ron Larson, Betsy FarberPublisher:PEARSONThe Basic Practice of StatisticsStatisticsISBN:9781319042578Author:David S. Moore, William I. Notz, Michael A. FlignerPublisher:W. H. FreemanIntroduction to the Practice of StatisticsStatisticsISBN:9781319013387Author:David S. Moore, George P. McCabe, Bruce A. CraigPublisher:W. H. Freeman