10. The owner of a company has asked you to conduct an evaluation of the customer...
Consumer Reports uses a 100-point customer satisfaction score to rate the nation's major chain stores. Assume that from past experience with the satisfaction rating score, a population standard deviation of o=12 is expected. Costco with its 432 warehouses in 40 states was the only chain store to earn an outstanding rating for overall quality. A sample of 15 Costco customer satisfaction scores is contained in the Excel Online file below. Construct a spreadsheet to answer the following questions. X Open...
Hypothesis Testing Method A = ADKAR Framework (column A) Method B = Prosci Change Management Methodology (column B) Null hypothesis is H0: Method A = Method B Research (Alternative) hypothesis is H1:Method A < Method B Sample size: 30 One-tailed test: Direction stated in the hypothesis is determine more effective (greater or less than). Level of Significance (a): .05 t-test is used because standard deviation of population is unknown, sample size is less than 30 Cutoff Sample Score (critical value)...
Consider a situation where we want to compare means, M1 and 42 of two populations, Group 1 and Group 2, respectively. A random sample of 40 observations was selected from each of the two populations. The following table shows the two-sample t test results at a = 5% assuming equal population variances: t-Test: Two-Sample Assuming Equal Variances Group 2 28652 33.460 40 Mean Variance Observations Pooled Variance Hypothesized Mean Difference d t Stat PTcut) one-tail Critical one-tail PTC-t) two-tail Critical...
3. Testing a population mean The test statistic (Chapter 11) Aa Aa You conduct a hypothesis test about a population mean u with the following null and alternative hypotheses: Ho: u-25.8 H1: <25.8 Suppose that the population standard deviation has a known value of a observations, which provides a sample mean of % 30.7. 17.8. You obtain a sample of n =62 Since the sample size large enough, you assume that the sample mean X follows a normal distribution. Let...
You conduct a hypothesis test about a population proportion p at a significance level of a = .01 using a random sample of size n = 38. Your test statistic follows a standard normal distribution when the null hypothesis is true as an equality, and its value obtained from the sample is z = -2.75. Use the Distributions tool to help you answer the questions that follow. Select a Distribution Distributions 0 1 2 3 If you perform a lower...
The sample data consist of 23 houses from a specific city yielded the average house price $226,460 and the standard deviation of the house price $11,500. Use a significance level 0.01 to test whether the mean house price of the whole city is more than $220,000. Compute the value of the test statistic, and P-value for the specified hypothesis test and state your conclusion. Assume the house prices of this city follows normal distribution. Question 2 options: Test statistic: t...
The sample data consist of 23 houses from a specific city yielded the average house price $226,460 and the standard deviation of the house price $11,500. Use a significance level 0.01 to test whether the mean house price of the whole city is more than $220,000. Compute the value of the test statistic, and P-value for the specified hypothesis test and state your conclusion. Assume the house prices of this city follows normal distribution. Test statistic: t = 2.69, p-value...
You are asked to conduct a simulation and run a hypothesis test at a significance level of 0.10. Based on your simulation, you find that the p-value is 0.051. What does this mean for your hypothesis test? 1. Since the p-value is greater than the significance level, you must fail to reject the null hypothesis. 2. Since the p-value is less than the significance level, you must fail to reject the null hypothesis. 3. Since the p-value is less than...
The Student's t distribution table gives critical values for the Student's t distribution. Use an appropriate d.f. as the row header. For a right-tailed test, the column header is the value of α found in the one-tail area row. For a left-tailed test, the column header is the value of α found in the one-tail area row, but you must change the sign of the critical value t to −t. For a two-tailed test, the column header is the value...
CAN YOU PLEASE FIX MY LAST TWO SENTENCES WITH THIS INFORMATION? You note in your report both the t critical for a one tailed and a two tailed test. Identify whether you need to use a one tailed or a two tailed test for the test statistic and t critical. Then only compare the test statistic with that critical value. Otherwise when you mention both, it looks like you don't know which one to use. t-Test: Two-Sample Assuming Unequal Variances...