A random sample of 100 observations from a population with standard deviation 18.99 yielded a sample mean of 93.4.
1. Given that the null hypothesis is μ=90 and the alternative
hypothesis is ?>90μ>90 using ?=.05α=.05, find the
following:
(a) Test statistic ==
(b) P - value:
(c) The conclusion for this test is:
A. Reject the null hypothesis
B. There is insufficient evidence to reject the
null hypothesis
C. None of the above
2. Given that the null hypothesis is μ=90 and the alternative
hypothesis is ?≠90 using ?=.05, find the following:
(a) Test statistic ==
(b) P - value:
(c) The conclusion for this test is:
A. Reject the null hypothesis
B. There is insufficient evidence to reject the
null hypothesis
C. None of the above
A random sample of 100 observations from a population with standard deviation 18.99 yielded a sample...
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...
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. ( , )...
(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.
A random sample of 100 observations from a population with standard deviation 76 yielded a sample mean of 114. Complete parts a through c below. a. Test the null hypothesis that u = 100 against the alternative hypothesis that u > 100, using a = 0.05. Interpret the results of the test. What is the value of the test statistic? und to two decimal places as needed.) Find the p-value. p-value = (Round to three decimal places as needed.) State...
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(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...
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(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 j = 90 and the alternative hypothesis is #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...
A random sample of 100 observations from a population with standard deviation 63 yielded a sample mean of 111. Complete parts a through c. a. Test the null hypothesis that y 100 against the alternative hypothesis that > 100, using a 0.05. Interpret the results of the test. Ho is rejected Ho is not rejected O Interpret the results of the test. Choose the correct interpretation below. O A. There is sufficient evidence to indicate the true population mean is...
A sample of size 100, taken from a population whose standard deviation is known to be 8.90, has a sample mean of 51.16. Suppose that we have adopted the null hypothesis that the actual population mean is greater than or equal to 52, that is, H0 is that μ ≥ 52 and we want to test the alternative hypothesis, H1, that μ < 52, with level of significance α = 0.05. a) What type of test would be appropriate in...