Find the minimum sample size required to obtain a level-a test such that B(Ha) = B for a two sided test on mean
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- a consumer advocay group tests the mean vitamin c content of 50 different brands of bottled juices using, in each case, a t-test of significance in which the null hypothesis is the mean amount of vitamin c that is on the nutrition facts label for that brand of juice. they find that two of the juice brands have statistically signifanctly lower vitamin c than claimed at the a 0.05 level. Is this an important discovery? Explain.For the Kruskal-Wallis ANOVA: Is there a statistical difference between the rankings of the core temperatures of the groups exposed to either a cold, neutral, or hot environment? (smaller number= coldest core temperature ranking, largest number= hottest core temperature ranking).” Ranks that are tied, get an average value of the two ranks. The Data is attached. Thank you.Suppose that a random sample of n = 5 was selected from the vineyard properties for sale in Sonoma County, California, in each of three years. The following data are consistent with summary information on price per acre for disease-resistant grape vineyards in Sonoma County. In Excel, carry out an ANOVA to determine whether there is evidence to support the claim that the mean price per acre for vineyard land in Sonoma County was not the same for each of the three years considered. Test at the 0.05 level and at the 0.01 level. 1996: 30000 34000 36000 38000 400001997: 30000 35000 37000 38000 400001998: 40000 41000 43000 44000 50000
- Test the research hypothesis that there is a difference in weights between the two groups at the 0.05 significance level? Assume the equality of variance, adequate sample size, and normal distribution for both the groups to test the hypothesis. X1 = 200 X2 = 198 n1 = 17 n2 = 17 s1 = 2.45 s2 = 2.35 Please show all the necessary steps of hypothesis testing to answer this question.A solar power engineer took a random sample of houses and installed the same type of solar panels using two different methods on each house to investigate whether there is a mean difference in the angles of installation between the two methods for all houses in the population of interest. The engineer found the sample mean difference between the two methods to be 0.2 degree and the p-value for a two-sided matched-pairs t-test for the mean difference to be 0.65. Assuming the conditions for inference are met, which of the following statements is the best interpretation of the p -value? (A) The probability that the null hypothesis is true is 0.65. (B) If the null hypothesis is true, the probability is 0.65 of observing a mean difference of 0.2 degree or -0.2 (C) (D) (E) degree. If the null hypothesis is true, the probability is 0.65 of observing a mean difference of greater than 0.2 degree or less than -0.2 degree. If the null hypothesis is true, the probability is 0.65 of observing a…In a test of weight loss programs, 111 subjects were divided such that 37 subjects followed each of 3 diets. Each was weighed a year after starting the diet and the results are in the ANOVA table below. Use a 0.05 significance level to test the claim that the mean weight loss is the same for the different diets. Source of Variation Between Groups Within Groups Total P-value 25.38918 0.8127 0.446353 F crit 3.080387 df MS F 50.778 3373.977 3424.755 2 108 31.24053 110 Should the null hypothesis that all the diets have the same mean weight loss be rejected? O A. Yes, because the P-value is greater than the significance level. O B. No, because the P-value is greater than the significance level. OC. No, because the P-value is less than the significance level. O D. Yes, because the P-value is less than the significance level.
- II-Term Exam Il-Make-up > الوقت المتبقي 40: The aim of a study is to test the ratio of variances in the weight of two groups of rats being under a certain drugs A (group 1) and B (group 2). A random sample of 17 rats was selected from group 1and 11 rats from group 2, Then their weight were recorded. The results are shown in the following table sd Median Mean n 1.05 3.5 17 group 1 2.0 2.20 11 group 2 3.0 1.15 F19,16,0.025 =0.386 F15,12,0.05=0.404 F16.10,0.025 = 0.335 F19,16,0.05=0.451 F15,12,0.1=0.496 | F16,10,0.05 = 0.401 F16,19,0.025 0.371 F12,15,0.05 =0.382 F10,16,0.025 =0.286| F16,19,0.05=0.437 F12,15,0.1=0.475 F10,16,0.05=0.354 Construct a 95% confidence interval for the ratio of the variances. [ Write only the final result in the box below ]To test a null hypothesis the 5 % level of significance is chosen, if the test is a left tailed test then the size of the critical region on the right tail will be Select one: O a. 0.025 ОБ. 0.005 O c. NONE O d. 0.0You wish to test the following claim (Ha) at a significance level of a = 0.10. Ho:μ = 76.3 Ha: μ < 76.3 You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size n = 72 with a mean of M = 72.5 and a standard deviation of SD = 8.7. What is the critical value for this test? Use the TI-84 or online t or z-score calculator. (Round to three decimal places.) Critical value= What is the test statist: for this sample? (Round to three decimal places.) Test statistic = The test statistic is... in the critical region not in the critical region This test statistic leads to a decision to... reject the null fail to reject the null As such, the final conclusion is that... (Watch the video below for help with the conclusion.) The sample data support the claim that the population mean is less than 76.3. There is not sufficient sample evidence to support the claim that the population mean is less than 76.3.