The manager at Round Table Pizza Parlor claims that the number of pizzas sold each day is not the same for each day of the week. The manager takes a simple random sample of 434 records from last week and they are listed below. At the level of significance of 5%, is there statistical evidence to support the manager's claim? Day of the week Number of pizzas Sun Mon Tue Wed Thu Fri Sat 54 67 55 52 55 73 78
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- Let's examine the mean of the numbers 1, 2, 3, 4, 5, 6, 7, and 8 by drawing samples from these values, calculating the mean of each sample, and then considering the sampling distribution of the mean. To do this, suppose you perform an experiment in which you roll an eight-sided die two times (or equivalently, roll two eight-sided dice one time) and calculate the mean of your sample. Remember that your population is the numbers 1, 2, 3, 4, 5, 6, 7, and 8. The true mean (µ) of the numbers 1, 2, 3, 4, 5, 6, 7, and 8 is , and the true standard deviation (o) is The number of possible different samples (each of size n = 2) is the number of possibilities on the first roll (8) times the number of possibilities on the second roll (also 8), or 8(8) = 64. If you collected all of these possible samples, the mean of your sampling distribution of means (µM) would equal and the standard deviation of your sampling distribution of means (that is, the standard error or ɑm) would be The following chart…In 2008, as an advertising campaign, the Nabisco Company announced a "1200 Chips Challenge," claiming that their 18-ounce bag of Chips Ahoy cookies contained an average of at least 1200 individual chocolate chips. Dedicated statistics students at the Air Force Academy (no kidding, they actually did this) purchased some randomly selected bags of cookies and counted the chocolate chips. Data from their sample is given below. %3D 1219, 1214, 1087, 1200 , 1419,1121 , 1325 , 1345 , 1244, 1258 , 1356, 1132, 1191 , 1270, 1295, 1135 Find a 95% confidence interval for the population mean number of chips in a bag of Chips Ahoy Cookies. (Give final answers to 2 decimal place accuracy) ANSWER: It can be stated with 95% confidence that the true population mean number of chocolate chips in a bag of Chips Ahoy Cookies lies in the interval <μ < chocolate chips per bag Based on this confidence interval, is the Nabisco Company justified in making their claim? O yes, because the true population mean is…A contributor for the local newspaper is writing an article for the weekly fitness section. To prepare for the story, she conducts a study to compare the exercise habits of people who exercise in the morning to the exercise habits of people who work out in the afternoon or evening. She selects three different health centers from which to draw her samples. The 57 people she sampled who work out in the morning have a mean of 4.8 hours of exercise each week. The 56 people surveyed who exercise in the afternoon or evening have a mean of 4.2 hours of exercise each week. Assume that the weekly exercise times have a population standard deviation of 0.8 hours for people who exercise in the morning and 0.3 hours for people who exercise in the afternoon or evening. Let Population 1 be people who exercise in the morning and Population 2 be people who exercise in the afternoon or evening. Step 1 of 2 : Construct a 95% confidence interval for the true difference between the mean amounts…
- A television station wishes to study the relationship between viewership of its 11 p.m. news program and viewer age (18 years or less, 19 to 35, 36 to 54, 55 or older). A sample of 250 television viewers in each age group is randomly selected, and the number who watch the station’s 11 p.m. news is found for each sample. The results are given in the table below. Age Group Watch11 p.m. News? 18 or less 19 to 35 36 to 54 55 or Older Total Yes 37 46 59 83 225 No 213 204 191 167 775 Total 250 250 250 250 1,000 (a) Let p1, p2, p3, and p4 be the proportions of all viewers in each age group who watch the station’s 11 p.m. news. If these proportions are equal, then whether a viewer watches the station’s 11 p.m. news is independent of the viewer’s age group. Therefore, we can test the null hypothesis H0 that p1, p2, p3, and p4 are equal by carrying out a chi-square test for independence. Perform this test by setting α = .05. (Round your answer to 3 decimal places.)…An Internet retailer would like to investigate the relationship between the amount of time in minutes a purchaser spends on its Web site and the amount of money he or she spends on an order. The table to the right shows the data from a random sample ofThe General Social Survey is an annual survey given to a random selection of about 1500 adults in the United States. Among the many questions asked are "What is the highest level of education you've completed?" and "If you're employed full-time, how many hours do you spend working at your job during a typical week?" In a recent year, 1079 respondents answered both questions. The summary statistics are given in the chart below. (The sample data consist of the times, in hours per week, that were given by the respondents.) Sample Sample Sample size Groups mean variance Less than h.s. High school Bachelor's 258 40.9 101.8 263 41.8 97.5 270 42.2 94.4 Graduate 288 43.0 98.7 Send data to calculator Send data to Excel To decide if there are any differences in the mean hours per week worked by these different groups, we can perform a one-way, independent-samples ANOVA test. Such a test uses the following statistic. Variation between the samples F= Variation within the samples For the data from…
- A television station wishes to study the relationship between viewership of its 11 p.m. news program and viewer age (18 years or less, 19 to 35, 36 to 54, 55 or older). A sample of 250 television viewers in each age group is randomly selected, and the number who watch the station’s 11 p.m. news is found for each sample. The results are given in the table below. Age Group Watch11 p.m. News? 18 or less 19 to 35 36 to 54 55 or Older Total Yes 49 59 61 84 253 No 201 191 189 166 747 Total 250 250 250 250 1,000 (a) Let p1, p2, p3, and p4 be the proportions of all viewers in each age group who watch the station’s 11 p.m. news. If these proportions are equal, then whether a viewer watches the station’s 11 p.m. news is independent of the viewer’s age group. Therefore, we can test the null hypothesis H0 that p1, p2, p3, and p4 are equal by carrying out a chi-square test for independence. Perform this test by setting α = .05. (Round your answer to 3 decimal places.)…A random sample of 10 compact cars, 10 mid-size cars, and 10 luxury cars were selected. The time (in seconds) each of the randomly selected cars required to accelerate from 0 to 60 mph was looked up on the Autos.com website. The results are presented below. Car Type Time (in seconds) Required to Accelerate from 0 to 60 mph Compact 9.3 8.1 10.2 8.8 9.0 9.3 7.7 9.2 9.4 8.6 Mid-Size 6.9 5.7 8.3 7.7 8.6 5.9 6.1 8.8 6.1 7.2 Luxury 5.7 6.3 5.4 4.7 6.2 7.0 5.9 5.3 6.3 5.0 Conduct a hypothesis test using ! = 0.05 to determine whether the mean time required to accelerate from 0 to 60 mph is the same for compact, mid-size, and luxury cars. (No accounting or excel, I need help solving it.)The Monterey Bay Aquarium, founded in 1984, is situated on the beautiful coast of Monterey Bay in the historic Cannery Row district. In 1985, the aquarium began a survey program that involved randomly sampling visitors as they exit for the day. The survey included visitor demographic information, use of social media, and opinions on their aquarium visit. For each visitor sampled during 2013-2015, the distribution of the number of children in their group is given. Number of Children 0 1 2 3 or more 2013 1855 585 599 515 Year 2014 2015 1751 1998 636 591 601 483 506 289 To access the complete data set, click the link for your preferred software format: CSV Excel JMP Mac-Text Minitab14-18 Minitab18+ PC-Text R SPSS TI CrunchIt! In this exercise, we will answer the question of whether there a significant difference in the distribution of the number of children in the group over this three-year period, and if so, how the distribution has changed. SOLVE: Create a graph of the data using…
- The safety director of a large steel mill took samples at random from company records of minor work-related accidents and classified them according to the time the accident took place. Time Number of Accidents Time Number of Accidents 8 up to 9 a.m. 10 1 up to 2 p.m. 11 9 up to 10 a.m. 9 2 up to 3 p.m. 12 10 up to 11 a.m. 12 3 up to 4 p.m. 7 11 up to 12 p.m. 6 4 up to 5 p.m. 11 Click here for the Excel Data File Using the goodness-of-fit test and the 0.10 level of significance, determine whether the accidents are evenly distributed throughout the day. H0: The accidents are evenly distributed throughout the day.H1: The accidents are not evenly distributed throughout the day. a. State the decision rule, using the 0.10 significance level. (Round your answer to 3 decimal places.) b. Compute the value of chi-square. (Round your answer to 3 decimal places.) c. What is your decision regarding H0?A real estate agent wishes to determine whether tax assessors and real estate appraisers agree on the values of homes. A random sample of the two groups appraised 10 homes. Is there a significant difference in the values of the homes for each group? Assume the data are from normally distributed populations. Real Estate Appraisers Tax Assessors: X1 = $83,256; s1 = $3256; n1 = 10 X2 = $88,354; s2 = $2341; n2 = 10Does the type of instruction in a college statistics class (either lecture or self-paced) influence students’ performance, as measured by the number of quizzes successfully completed during the semester? Twelve students were recruited for the study. Students received the lecture instruction or a self-paced instruction. The total number of quizzes completed during teach type of instruction appear below.