Determine (a) the x2 test statistic, (b) the critical value using a = 0.05, and (c a = 0.05 level of significance. %3D 1 Ho: PA = PB = Pc = PD 4 H,: At least one of the proportions is c
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- A pharmaceutical company is running a test comparing two forms of advertising for the same product one which involved a television campaign and the other a print campaign. Market sectors are randomly assigned to receive a particular form of advertising and product sales are recorded during the month following ad campaigns. Using the data provided test if there is a significant difference in product sales by method of advertisement. Use (alpha) a=0.05After running a regression you conduct a hypothesis test of Ho: ß = 2.5 versus Hạ: B + 2.5 is performed using a = 0.10. The value of the test statistic z = -1.80 (the critical value is -1.645). If the true value of B is 2.5, does the conclusion you reach result in Type I, Type II error, or the correct decision? O Type II Error O Type I Error O correct decision not enough informationThe lowest level of significance to reject the null hypothesis of no linear association between blood pressure and age is: OA: 0.003 OB: 0.05 OC: 0.0002 OD: 0.0001 OE: 0.04
- Calculate the Pearson Correlation Coefficient for these two variables and b) test whether the correlation is significantly different from 0. Run the test at a 5% level of significance. Give 1) the appropriate null and alternative hypotheses; 2) the appropriate test; 3) the decision rule; 4) the calculation of the test statistic; and 5) your conclusion including a comparison to alpha or the critical value. ID Hours of Study (x) Exam Score (y) 1 2 80 2 3 83 3 5 95 4 1 79 5 2 84 6 3 90A study is being conducted to compare vitamin C and zinc to determine which is better at fighting colds. Customers believe vitamin C is better at fighting colds. What are the appropriate hypotheses for this testing scenario? Let μc equal the mean of the effectiveness of vitamin C and μz equal the mean of the effectiveness of zinc. O Ho: Hc - Hz= 0 Ha Hc - Hz > 0 O Ho: Hc - Hz= 0 Ha Mc-Hz 0 o Ho: Xe – xz = 0 Hai Xi — Xz #0The yield of a chemical process is being studied. The two most important variables are thought to be pressure and the temperature. Three levels of each factor are selected, and a factorial experiment with the two replicates is performed. The yield data follow: Question: (Use Alpha = 0.05) Is there any indication that either factor influences brightness? Do the two factors interact?
- Physical properties of six flame-retardant fabric samples were investigated in an article. Use the accompanying data and a 0.05 significance level to determine whether a linear relationship exists between stiffness x (mg-cm) and thickness y (mm). O Ho: P = 0 H₂: P = 0 O Ho: P = 0 H₂:p>0 State the appropriate null and alternative hypotheses. O Ho: P = 0 Ha: P < 0 O Ho: P = 0 Ha: P = 0 X y r = 7.95 24.63 12.39 7.06 24.17 35.70 0.25 0.64 0.32 0.27 0.80 0.56 Compute the value of the sample correlation coefficient, r. Round your answer to four decimal places. Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.) t = P-value = State the conclusion in the problem context. O Fail to reject Ho. The data does not indicate that the population correlation coefficient differs from 0. O Reject Ho. The data indicates that the population correlation coefficient differs from 0. O Reject Ho. The data does not…The data in the table below presents the hourly quantity of production for three lines of production processes over the first 4 days in XYZ Company. Answer the questions based on the Excel Output given in the picture below. State the null and alternative hypothesis for single factor ANOVA. State the decision rule (α = 0.05). Calculate the test statistic. Make a decision.A random sample of 15 families representing three social classes has been observed for the frequency with which the parents administer physical punishment to the children over a period of a week. Are the differences significant? Use α = .05 and the five-step model to conduct your hypothesis test. Write a sentence or two interpreting your results. Working Class: 10,9,4,2,1 Middle Class: 11,10,5,2,0 Upper Class: 7,5,2,0,0
- A random sample of senior citizens living in a retirement village reported that they had an average of 1.42 face-to-face interactions per day with their neighbors. A random sample of those living in age-integrated communities reported 1.58 interactions per day. Is the difference between the two senior citizen groups significant? Please provide t critical, t obtained and your decision. Retirement Community 1 Retirement Community 2 Mean1 = 1.42 Mean2 = 1.58 s1 = 0.10 s2 = 0.78 N1 = 43 N2 = 37Suppose IQ scores were obtained for 20 randomly selected sets of couples. The 20 pairs of measurements yield x = 101.16, y = 102.3, r= 0.810, P-value = 0.000, and y = 23.09+0.78x, where x represents the IQ score of the wife. Find the best %3D %3D %3D predicted value of y given that the wife has an IQ of 109? Use a significance level of 0.05. Click the icon to view the critical values of the Pearson correlation coefficient r. .... The best predicted value of y is . (Round to two decimal places as needed.) Critical values of the pearson correlation coefficient r NOTE: To test Ho: p = 0 against H,: p#0, reject Ho |if the absolute value of r is greater than the critical value in the table. a = 0.05 a = 0.01 4 0.950 0.990 0.878 0.959 0.811 0.917 0.754 0.875 0.707 0.834 0.666 0.798 10 0.632 0.765 11 0.602 0.735 12 0.576 0.708 13 0.553 0.684 14 0.532 0.661 15 0.514 0.641 16 0.497 0.623 17 0.482 0.606 18 0.468 0.590 19 0.456 0.575 20 0.444 0.561 25 0.396 0.505 30 0.361 0.463 35 0.335 0.430 40…(1) Two rubber compounds were tested for tensile strength and the fol- lowing values were found A : 32,30,33, 32, 29, 34, 32 B: 33,35, 36, 37,35, 34 Under the assumption that the two populations are normally distributed, test the hypothesis that the average tensile strength of the two rubber compounds is different using significance level a = 0.01 and a = 0.05.