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- For the population of college students, the number of hours spent studying per week is normally distributed with a = 4.40 and = 2.32 hours. We surveyed a random sample of students majoring in psychology about the number of hours they spent studying this past week. The data is as follows: 6, 8, 9, 4, 3, 11, 4, 6, 5, 7 What is the dependent variable? What is the level of measurement of the dependent variable? 3. Is there a significant increase in the number of hours spent studying for our sample compared to the population of college students? Perform a complete hypothesis test (z-test) using alpha = .01 level. SHOW ALL WORKA statistical program is recommended. The National Football League (NFL) records a variety of performance data for individuals and teams. To investigate the importance of passing on the percentage of games won by a team, the following data show the conference (Conf), average number of passing yards per attempt (Yds/Att), the number of interceptions thrown per attempt (Int/Att), and the percentage of games won (Win%) for a random sample of 16 NFL teams for one full season. Team Conf Yds/Att Int/Att Win% Arizona Cardinals NFC 6.5 0.042 50.0 Atlanta Falcons NFC 7.1 0.022 62.5 Carolina Panthers NFC 7.4 0.033 37.5 Cincinnati Bengals AFC 6.2 0.026 56.3 Detroit Lions NFC 7.2 0.024 62.5 Green Bay Packers NFC 8.9 0.014 93.8 Houstan Texans AFC 7.5 0.019 62.5 Indianapolis Colts AFC 5.6 0.026 12.5 Jacksonville Jaguars AFC 4.6 0.032 31.3 Minnesota Vikings NFC 5.8 0.033 18.8 New England Patriots AFC 8.3 0.020 81.3 New Orleans Saints NFC 8.1 0.021 81.3 Oakland Raiders AFC…A physical therapist wanted to know whether the mean step pulse of men was less than the mean step pulse of women. She randomly selected 54 men and 70 women to participate in the study. Each subject was required to step up and down a 6-inch platform. The pulse of each subject was then recorded. The following results were obtained. Two sample T for Men vs Women N Mean StDev SE Mean Men Women 98% CI for mu Men - mu Women (- 12.20, - 1.00) T-Test mu Men = mu Women (vs H2 O C. Ho: H1 = H2; Ha: H1 #H2 (b) Identify the P-value and state the researcher's conclusion if the level of significance was a = 0.01. What is the P-value? P-value =
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- The parameters of my population distribution are: µ = 20 and σ = 5. The parameters of my sample are: Mean = 24 and n = 25. Calculate my sample's Z score on the comparison distribution.In each situation, find the value of the t-statistic for the sample mean x and give the value of degrees of freedom (df). (a) x = 6, µ = 12, o = 24, n = 25 t = df = (b) x = 18, u = 12, o = 24, n = 25 t = df =a survey is conducted to compare the proportions of men and women whoaccess their bank statements online. Denoting the population propotions by Pm and Pw, a two-proportion z-test is carried out to test Ho: pm = pw against Ha: pm > pw. The values of the test statistic is found to be z = 0.784, and the p-value for the test is found to be 0.216. Which of the following is the correct interpretation of the p-value? a) given that ha is true, the probability of getting a value of z at least as large as 0.784 is 0.216. b) given that ho is true, the probability of getting a value of z at least as large as 0.784 is 0.216. c) given the results of the survey, the probability that Pm > Pw is 0.216. d) given the results of the survey, the probability that Pm = Pw is 0.216. e) none of the above
- A one-way ANOVA was conducted at ? = 0.05. The study involved three (3) different groups with 12 subjects in each. The means for each group were as follows: ?1 = 10.92, ?2 = 13.83, ??? ?3 = 14.92. The sum of squares values is shown in the table below.You are conducting a study to see if the proportion of voters who prefer Candidate A is significantly more than 0.46. Your sample data produce the test statistic z=1.883. Find the p-value accurate to 4 decimal places.A standardized exam’s score are normally distributed. In a recent year, the mean test score was 1477 and the standard was 315. The test scores of four students selected at random 1860 1230, 2160, and 1360. Find the Z- score that correspond to each value and determine wether any of the values are unusual. The Z-score for 1860 is-The Z-score for 1230 is-The Z-score for 2160 is-The Z-score for 1360is-
