Question

In: Statistics and Probability

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 than​100, 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 the test.

c. compare the results of the two tests you conducted. explain why the results differ.

Solutions

Expert Solution


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 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 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 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...
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 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...
(1 point) A random sample of 100100 observations from a population with standard deviation 19.788150778587319.7881507785873 yielded...
(1 point) A random sample of 100100 observations from a population with standard deviation 19.788150778587319.7881507785873 yielded a sample mean of 93.893.8. (a)    Given that the null hypothesis is ?=90μ=90 and the alternative hypothesis is ?>90μ>90 using ?=.05α=.05, find the following: (i)    critical z/t score     equation editor Equation Editor (ii)    test statistic == (b)    Given that the null hypothesis is ?=90μ=90 and the alternative hypothesis is ?≠90μ≠90 using ?=.05α=.05, find the following: (i)    the positive critical z/t score     (ii)    the negative critical z/t score     (iii)    test statistic ==...
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 =
A random sample is drawn from a population with mean μ = 70 and standard deviation...
A random sample is drawn from a population with mean μ = 70 and standard deviation σ = 5.8. [You may find it useful to reference the z table.] a. Is the sampling distribution of the sample mean with n = 17 and n = 43 normally distributed? Yes, both the sample means will have a normal distribution. No, both the sample means will not have a normal distribution. No, only the sample mean with n = 17 will have...
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
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT