(1 point) A random sample of 100
observations from a population with standard deviation 25.11 yielded a sample mean of 94.2
.
1. Given that the null hypothesis is
(1 point) A random sample of 100 observations from a population with standard deviation 25.11 yielded a sample mean of 9...
(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 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 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...
(1 point) The sample mean and standard deviation from a random sample of 33 observations from a normal population were computed as x¯=34 and s = 10. Calculate the t statistic of the test required to determine whether there is enough evidence to infer at the 6% significance level that the population mean is greater than 30. Test Statistic =
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...
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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 n = 100 observations is selected from a population with mean 20 and standard deviation 15. What is the probability of observing a mean greater than 21?
(1 point) A random sample of 100 observations produced a mean of 2 = 32 from a population with a normal distribution and a standard deviation 4.1. (a) Find a 95% confidence interval for a 25.33 SAS 26.47 (b) Find a 90% confidence interval for 25.48 SHS 26.37 (c) Find a 99% confidence interval for 25.15 SHS 26.65
A random sample of 100 observations is drawn from a population with a mean equal to 20 and standard deviation equal to 16. Let Xbar denote the sample mean. (a) Find the mean and standard deviation of the sampling distribution of Xbar (b) Describe the shape of the sampling distribution of Xbar. Does your answer depend on the sample size? (c) Find P(Xbar <16). P(Xbar >23). P(16< Xbar <23). Thank You!