A random sample of 100 observations from a population with standard deviation 23.99 yielded a sample...

A random sample of 100 observations from a population with standard deviation 23.99 yielded a sample mean of 94.1

1. Given that the null hypothesis is μ=90μ=90 and the alternative hypothesis is μ>90μ>90 using α=.05α=.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 the null hypothesis
C. None of the above

2. Given that the null hypothesis is μ=90μ=90 and the alternative hypothesis is μ≠90μ≠90 using α=.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

Solutions

Expert Solution

orchestra answered 1 year ago

Next > < Previous

Related Solutions

a random sample of 100 observations from a population. with standard deviation 60 yielded a sample...

a random sample of 100 observations from a population. with standard deviation 60 yielded a sample mean of 110. A. test the null hypothesis that m = 100 and the alternative hypothesis m > 100 using alpha = .05 and interpret the results b. test the null against the alternative hypothesis that m isn't equal to 110. using alpha = .05 and interpret the results c. compare the p values of the two tests you conducted. Explain why the results...

a random sample of 100 observations from a population. with standard deviation 60 yielded a sample...

A random sample of 100 observations from a population with standard deviation 17.83 yielded a sample...

A random sample of 100 observations from a population with standard deviation 17.83 yielded a sample mean of 93. 1. Given that the null hypothesis is ?=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. There is insufficient evidence to reject the null hypothesis B. Reject the null hypothesis C. None of the above 2. Given that the null hypothesis is...

A random sample of 100 observations from a population with standard deviation 14.29 yielded a sample...

A random sample of 100 observations from a population with standard deviation 14.29 yielded a sample mean of 92.7. 1. Given that the null hypothesis is μ=90 and the alternative hypothesis is μ>90 using α=.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 the null hypothesis C. None of the above 2. Given that the null hypothesis is...

A random sample of 100 observations from a population with standard deviation 70 yielded a sample...

A random sample of 100 observations from a population with standard deviation 70 yielded a sample mean of 113. Complete parts a through c below. a. Test the null hypothesis that muequals100 against the alternative hypothesis that mugreater than100, using alphaequals0.05. Interpret the results of the test. What is the value of the test statistic? b. test the null hypothesis that mu = 100 against the alternative hypothesis that mu does not equal 100, using alpha=.05. interpret the results of...

A random sample of 100 observations from a population with standard deviation 22.99 yielded a sample...

A random sample of 100 observations from a population with standard deviation 22.99 yielded a sample mean of 94.1. 1. Given that the null hypothesis is μ≤90 and the alternative hypothesis is μ>90 using α=.05, find the following: (a) Test statistic = (b) P - value: (c) The decision for this test is: A. Fail to reject the null hypothesis B. Reject the null hypothesis C. None of the above 2. Given that the null hypothesis is μ=90 and the...

A random sample of n observations is selected from a population with standard deviation σ =...

A random sample of n observations is selected from a population with standard deviation σ = 1. Calculate the standard error of the mean (SE) for these values of n. (Round your answers to three decimal places.) (a) n = 1 SE = (b) n = 2 SE = (c) n = 4 SE = (d) n = 9 SE = (e) n = 16 SE = (f) n = 25 SE = (g) n = 100 SE =

In a random sample of 100 measurements from a population with known standard deviation 200, the...

In a random sample of 100 measurements from a population with known standard deviation 200, the average was found to be 50. A 95% confidence interval for the true mean is

A sample of 61 observations yielded a sample standard deviation of 5. If we want to...

A sample of 61 observations yielded a sample standard deviation of 5. If we want to test Ho: σ²=20, the test statistic is:

A statistics practitioner took a random sample of 52 observations from a population whose standard deviation...

A statistics practitioner took a random sample of 52 observations from a population whose standard deviation is 27 and computed the sample mean to be 100. Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits. A. Estimate the population mean with 95% confidence. Confidence Interval = B. Estimate the population mean with 95% confidence, changing the population standard deviation to 45; Confidence Interval =...

Subjects