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- A sleep therapist wanted to see if a herbal tea advertised as a sleep aid really worked. He located 46 people with sleep problems and matched them into pairs on the basis of sleep problems and suggestibility. He then randomly assigned one person from each pair to drink the tea at bedtime (the experimental group), while the control group went to sleep as they normally did. He used an EEG to measure the minutes to sleep onset (the fewer the minutes to sleep onset, the better). He found MControl = 21.20, MExperimental = 19.70, sD = 5.47. Use an alpha of .05 and a two-tailed test to determine if the herbal teach was effective.MAKE SURE YOU SELECT 2 ANSWERS FOR THIS QUESTION - Type of test and hypothesis. Professor Diff E. Cult wonders if male and female students earn significantly different grades when taking the same course with the same instructor. He randomly samples 160 males and 100 females from selected classes. The males receive 20 As, 32 Bs, 70 Cs, 30 Ds, and 8 Fs. The females receive 20 As, 28 Bs, 30 Cs, 10 Ds, and 12 Fs. His hypothesis is that gender and grades are related. Choose the type of test you should use to test this hypothesis? Then choose the correct hypotheses in symbol notation (or words if symbol is not appropriate). O z-test single sample t-test dependent t-test O independent t-test Pearson correlation t-test ANOVA O Chi-Square Goodness of Fit Test Chi-Square Test of Independence O Ho: H1 - P2 = 0 H1: P1 - P2 # 0 O Ho: µ1 - P2 < 0In 2016, the New York Times reported that the average American spent 50 minutes per day on Facebook, as compared to an average of 42 minutes per day reading, exercising, or actually socializing combined. You suspect that this gap has widened over the years, and collect a random sample of 100 people. For 50 of them (again, randomly chosen from your sample), you track their time spent on Facebook; you track the other 50 peoples’ total time spent reading/exercising/socializing. (a) Say the 2016 figures regarding Americans’ time allotments still hold true. If the population standard deviation for minutes using Facebook is 12.5 minutes, and the population standard deviation for minutes spent on reading, exercising, and socializing (referred to from here on as ”Other” activities) is 8 minutes, what are the expected value and the standard error of the difference between the average amount of time spent on Facebook and the average amount of time spent on Other activities computed from a sample…
- Two different blood pressure medicines are being compared to determine if the average reduction in blood pressure is the same for each medication. The goal of the study is to determine if the medications differ. Twenty men age 50-60 years old are selected for the study. Ten men are chosen at random to receive the first medication and the other 10 men receive the second medication. Each of the 20 men is monitored for one month to determine the change in blood pressure over that time. Minitab provides the 95% confidence interval for (mu1 - mu2) (2.63, 14.18) a. Interpret this 95% CI. b. What assumptions (be specific) are necessary to construct this CI?THE MANAGER OF A SERVICE STATION IS IN THE PROCESS OF ANALYZING THE NUMBER OF TIMES CAR OWNERS CHANGE THE OIL IN THEIR CARS. SHE BELIEVES THAT THE AVERAGE MOTORIST CHANGES HIS OR HER CAR'S OIL LESS FREQUENTLY THAN RECOMMENDED BY THE OWNER'S MANUAL (TWO TIMES PER YEAR). IN A PRELIMINARY SURVEY SHE ASKED 14 CAR OWNERS HOW MANY TIMES THEY CHANGED THEIR CARS' OIL IN THE LAST 12 MONTHS. THE RESULTS ARE 1, 1, 2, 0, 3, 3,0, 1, 0, 1, 2, 3, 3, AND 1. 2.1 Does this data provide sufficient evidence at the 10% significance level to indicate that the manager is correct? 2.2 Ho: H₁: Rejection region: Test statistic: Decision: Conclusion: What condition is required in order to analyze this data using a t-test?I need help answering this question
- please explain!Sophomore, junior, and senior students at a high school will be surveyed regarding a potential increase in the extracurricular student activities fee. There are three possible responses to the survey question - agree with the increase, do not agree with the increase, or no opinion. A chi-square test will be conducted to determine whether the response to this question is independent of the class in which the student is a member. How many degrees of freedom should the chi-square test have?On average, a sample of n = 36 scores will provide a better estimate of the population mean than a sample of n = 49 scores from the same population.
- There exists 2 treatments for a specific illness. One uses pills, and the other infrared pulses. To measure the illness, you measure the level of the a particular marker in the blood; the higher the marker, the worse the disease. Which of the following tests the claim that the infrared pulse treatment improves outcomes better ?A personal trainer wanted to test the effectiveness of two different workout routines. She took a random sample of 70 of her clients and, in a random order, had them complete one of the two routines for two weeks. Then, after a one-month waiting period, she asked them to come back and do the other routine for two weeks. After each two-week period of the exercise routines, she measured their performance on a physical fitness aptitude test. She found the average difference in aptitude scores between the two routines for each client was 30.4 with a standard error of 3.2. What would be her 85% confidence interval for the average difference in the effectiveness of the two routines? (3 decimal places) ( , )Some people seem to believe that you can fix anything with duct tape. Even so, many were skeptical when researchers announced that duct tape may be a more effective and less painful alternative than liquid nitrogen, which doctors routinely use to freeze warts. An article described a study conducted at a medical center. Patients with warts were randomly assigned to either the duct tape treatment or the more traditional freezing treatment. Those in the duct tape group wore duct tape over the wart for 6 days, then removed the tape, soaked the area in water, and used an emery board to scrape the area. This process was repeated for a maximum of 2 months or until the wart was gone. Data consistent with values in the article are summarized in the following table. Number with Wart Treatment Successfully Removed Liquid nitrogen freezing 103 60 Duct Tape 100 88 A USE SALT Do these data suggest that freezing is less successful than duct tape in removing warts? Test the relevant hypotheses using a…