thank you for all the help! i appreciate everythjng! 1 2 3 (1 point) A random...
im so grateful for all the help! thank you in advance !! this means so much to me you havw no idea 1 2 3 4 At least one of the answers above is NOT correct. (1 point) (a) Find the P-value for the test statistic z = -1.32 for the following null and alternative hypotheses: Ho: The population mean is 18. H : The population mean is less than 18. The P-value is 0.000347 (b) Find the P-value for...
thank you for help! it means a lot to me ! and can u please underline or circle the answers. thank you 1 2 (1 point) A random sample of 100 observations from a population with standard deviation 13.83 yielded a sample mean of 92.3. Given that the null hypothesis is u = 90 and the alternative hypothesis is u > 90 using a = .05, find the following: (a) Test statistic: (b) P-value: DOO The conclusion for this test...
please help. thank you so much !! 1 2 0.0702 0.0702 incorrect incorrect At least one of the answers above is NOT correct. 1 of the questions remains unanswered. (1 point) A random sample of 100 observations from a population with standard deviation 8.31 yielded a sample mean of 91.5. Part 1: Part 2: - Part 3 Given that the null hypothesis is p = 90 and the alternative hypothesis is x 90 using a = .05, find the following:...
Ch6 Sec2: Problem 3 PreviousProblem List Next (1 point) A random sample of 100 observations from a population with standard deviation 9.13 yielded a sample mean of 91.6 1. Given that the null hypothesis is--90 and the alternative hypothesis is μ > 90 using a) Test statistic- (b) P- value: (c) The conclusion for this test is: 05, find the following: A. Reject the null hypothesis B. There is insufficient evidence to reject the null hypothesis C. None of the...
(1 point) A random sample of 100 observations from a population with standard deviation 5.24 yielded a sample mean of 91. Part 1: Given that the null hypothesis is j = 90 and the alternative hypothesis is ju > 90 using : .05, find the following: (a) Test statistic = A = (b) P-value: Part 2
(1 point) A random sample of 100 observations from a population with standard deviation 25.72 yielded a sample mean of 94.3. Part 1: Given that the null hypothesis is u = 90 and the alternative hypothesis is j > 90 using a = .05, find the following: (a) Test statistic = (b) P-value: Note: Round off the test statistic to 2 decimal places and the P-value to 4 decimal places.
Homework4: Problem 2 Previous Problem Problem List Next Problem (6 points) A random sample of 100 observations from a population with standard deviation 23.75 yielded a sample mean of 94.5. 1. Given that the null hypothesis is H = 90 and the alternative hypothesis is u > 90 using a = .05, find the following: (a) Test statistic = (b) P-value: (c) The conclusion for this test is: A. There is insufficient evidence to reject the null hypothesis B. Reject...
In a sample of credit card holders the mean monthly value of credit card purchases was $ 400 and the sample variance was 67 ($ squared). Assume that the population distribution is normal. Answer the following, rounding your answers to two decimal places where appropriate. (a) Suppose the sample results were obtained from a random sample of 12 credit card holders. Find a 95% confidence interval for the mean monthly value of credit card purchases of all card holders. ( , )...
Hello, I really need help on problems 1-4 listed below for my practice set. I appreciate the help! 1. Which of the following is true with regard to a P-value? (More than one answer may be correct) a. The P-value is generated under the assumption that the null hypothesis is false. b. The P- value may never be negative. c. As the P-Value approaches 0 is it is less likely the null hypothesis will be rejected. d. In a left-tailed...
THANK YOU IN ADVANCE FOR HELPING ME! i appreciate this soo much! this has helped me a lot especially throught times like this. im so grateful for your help 1 2 3 4 (1 point) The time needed for college students to complete a certain paper-and-pencil maze follows a normal distribution with a mean of 30 seconds and a standard deviation of 2.3 seconds. You wish to see if the mean time is changed by vigorous exercise, so you have...