28. What is the appropriate null hypothesis for this test? A. Ho: u =H2 3DH3 3D H4 B. Ho: µ1 = H23D43 C. Ho: H1 = Hz С.
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- Consider a study using a between-groups design with between-groups df = 3 and within-groups df = 4. Given an F ratio of 6.8, the researcher should: a. reject the null hypothesis if alpha is .05 but fail to reject if alpha of .01 b. reject the null hypothesis if alpha is .01 but fail to reject if alpha of .05 c. reject the null for both alpha = .01 or alpha = .05 d. fail to reject the null hypothesis whether alpha is .01 or .05a researcher examined a number of monthly meetings (2 or 4) on job satisfaction. if the t obtained for a correlated groups t-test was 2.50, df=3, and the t critical = 3.182, what s/he conclude?A researcher is using a two-tailed hypothesis test with α = 0.01 to evaluate the effect of a treatment. If theboundaries for the critical region are t = ± 2.845, then how many individuals are in the sample?A. n = 23B. n = 22C. n = 21D. n = 20E. cannot be determined from the information given
- For conducting a two-tailed hypothesis test with a certain data set, using the smaller of n1-1 and n2-1 for the degrees of freedom results in df=11, and the corresponding critical values are t=+-2.201. Using the formula for the exact degrees of freedom results in df=19.063, and the corresponding critical values are t=+-2.093. How is using the critical values of t=+-2.201 more "conservative" than using the critical values of +- 2.093?A study to compare two insurance companies on length of stay for pediatric asthma patients randomly sampled 393 cases from Insurer A. An independent random sample of 396 cases from Insurer B gave the results on length of stay, and the summary statistics are show below. Insurer A (Group 1): x⎯⎯⎯x¯ = 2.31, ss = 1.22, nn = 393Insurer B (Group 2): x⎯⎯⎯x¯ = 2.96, ss = 1.5, nn = 396 Unless otherwise stated, give your answers to three decimal places. Construct a 99% confidence interval for the difference in average length of stay between Insurer A and Insurer B. Give your answers to three decimal places. Use t∗=2.582t∗=2.582.( , ) Conduct a hypothesis test to determine whether the average length of stay for Insurer A is different from than the average length of stay for Insurer B. What is the parameter of interest?i. ppii. μμiii. p1−p2p1−p2iv. μ1−μ2μ1−μ2v. μdμd What is the correct null value for this test? What sign should appear in the alternative hypothesis?i. == ii.…You wish to test the following claim (HaHa) at a significance level of α=0.002α=0.002. Ho:p1=p2Ho:p1=p2 Ha:p1≠p2Ha:p1≠p2 You obtain 480 successes in a sample of size n1=647n1=647 from the first population. You obtain 562 successes in a sample of size n2=737n2=737 from the second population. For this test, you should NOT use the continuity correction, and you should use the normal distribution as an approximation for the binomial distribution. What is the z-score of the critical value? (Report answer accurate to three decimal places.)±± What is the standardized test statistic for this sample? (Report answer accurate to three decimal places.) The test statistic is... in the critical region not in the critical region This test statistic leads to a decision to... reject the null accept the null fail to reject the null
- For a hypothesis test of H0:μ=μ0H0:μ=μ0 against the alternative Ha:μ<μ0Ha:μ<μ0, the z-test statistics is found to be 2.00. This finding is a.significant at neither the 0.01 nor the 0.05 levels. b.significant at the 0.01 level but not at the 0.05 level. c.significant at both the 0.01 and the 0.05 levels. d.not large enough to be considered significant. e.significant at the 0.05 level but not at the 0.01 level.A group of Sports Science students (n = 20) are selected from the population to investigate whether a 12-week plyometric-training programmed improves their standing long jump performance. In order to test whether this training improves performance, the students are tested for their long jump performance before they undertake a plyometric-training programme and then again at the end of the programmed .The following table present the results of before and after the training programmed. Test the claim that their long jump performance is higher after the training. Student Number (Before the training) Jump1 (After the training) Jump2 1 2.25 2.24 2 2.42 2.48 3 2.26 2.29 4 2.58 2.62 5 2.62 2.64 6 2.16 2.18 7 2.40 2.44 8 2.62 2.67 9 2.35 2.39 10 2.44 2.47 Follow the steps in hypothesis testing. Answer the following. What is the appropriate statistical test to use?…Correlated t tests generally have more power than independent-groups t tests because the former usually have smaller standard errors of the difference between means, making the t value bigger. True False
- You wish to test the following claim (HaHa) at a significance level of α=0.005α=0.005. a) Ho:p1=p2Ho:p1=p2 b) Ha:p1<p2Ha:p1<p2You obtain 222 successes in a sample of size n1=404n1=404 from the first population. You obtain 388 successes in a sample of size n2=623n2=623 from the second population. For this test, you should NOT use the continuity correction, and you should use the normal distribution as an approximation for the binomial distribution.What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = _____What is the p-value for this sample? (Report answer accurate to four decimal places.)p-value = ______An employee of a bank wants to test the null hypoth-esis that on the average the bank cashes 10 bad checks per day against the alternative that this figure is too small. Ifhe takes a random sample and decides to reject the nullhypothesis if and only if the mean of the sample exceeds12.5, what decision will he make if he gets x = 11.2, andwill it be in error if(a) λ = 11.5; (b) λ = 10.0?Here λ is the mean of the Poisson population beingsampled.A random sample of 80 poultry farms of one variety gave an average production of 240 eggs per bird. Another random sample of 50 poultry farms of another variety gave an average production of 195 eggs per bird. At a=0.05 is there sufficient evidence to support that there is a significant difference between the egg production of the two varieties of birds. Assume o1 =18 eggs and o2=15 eggs
