76 subjects were placed on a low-fat diet. After 12 months, their mean weight loss was...

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76 subjects were placed on a low-fat diet. After 12 months, their mean weight loss was 2.2 kg. The population standard deviation is assumed to be 6.1 kg. Can we conclude that the mean weight loss is greater than 0, making it effective? If we use a = .05, select all of the following that are true. If we used a = .02, the critical value would be greater than ours used at a =.05. We do conclude that the mean weight loss is greater than 0. In computing the p-value, we need to multiply our normalcdf answer by two. In computing the test statistic value, the standard error of the mean, to three decimal accuracy, is .700. When using the classical approach, the observed value of the sample mean lies to the right of the critical value. The Central Limit Theorem is needed to complete this problem. J We would reach the opposite conclusion, if we used .01 as our level of significance. We assume at the start that = 0. When using the test statistic approach, the critical value (to three decimal accuracy) is 1.960. When using the classical approach, the critical value is 1.15, to two decimal accuracy. 76 subjects were placed on a low-fat diet. After 12 months, their mean weight loss was 2.2 kg. The population standard deviation is assumed to be 6.1 kg. Can we conclude that the mean weight loss is greater than 0, making it effective? If we use a = .05, select all of the following that are true. If we used a = .02, the critical value would be greater than ours used at a =.05. We do conclude that the mean weight loss is greater than 0. In computing the p-value, we need to multiply our normalcdf answer by two. In computing the test statistic value, the standard error of the mean, to three decimal accuracy, is .700. When using the classical approach, the observed value of the sample mean lies to the right of the critical value. The Central Limit Theorem is needed to complete this problem. J We would reach the opposite conclusion, if we used .01 as our level of significance. We assume at the start that = 0. When using the test statistic approach, the critical value (to three decimal accuracy) is 1.960. When using the classical approach, the critical value is 1.15, to two decimal accuracy.