TOPIC 8 Chi-Square goodness-of-fit test Problem 12.1 Use a chi-square goodness-of-fit to determine whether the observed frequencies are distributed the same as the expected frequencies (α = .05) Category | fo | fe | 1 | 53 | 68 | 2 | 37 | 42 | 3 | 32 | 33 | 4 | 28 | 22 | 5 | 18 | 10 | 6 | 15 | 8 | Step 1 Ho: The observed frequencies are distributed the same as the expected frequencies Ha: The observed frequencies are not distributed the same as the expected frequencies Step 2 df = k – m – 1 Step 3 α = 0.05 x 2 0.05, 5df = 11.0705 Step 4 Reject Ho if x 2 > 11.0705 Category | fo | fe | | 1 | 53 | 68 | | 2 | 37 | 42 | | 3 | 32 | 33 | | 4 | 28 | 22 | | 5 | 18 | 10 | | 6 | 15 | 8
FASB states that if some amount within a range of loss appears at the time to be a better estimate than any other amount within the range, that amount shall be accrued (FASB 450-20-30-1). Therefore, M should record a liability in the amount of $17 million. 2. For the year- end December 31, 2009, financial statements, should M adjust its liability? If so, what amount should be recorded; and should the amount of the adjustment be considered a 2009 event or a prior period adjustment?
What value is stored in uninitialized variables? An uninitialized variable is by default a 0. Algorithm Workbench 3-10 3. Write assignment statements that perform the following operations with the variable a, b and c. a) Adds 2 to a and stores the result in b. { b = 2 + a } b) Multiplies b by 4 and stores the result in a.
Answer: False Difficulty: Medium 3. The sampling distribution of [pic] must be a normal distribution with a mean 0 and standard deviation 1. Answer: False Difficulty: Medium (REF) 4. For any sampled population, the population of all sample means is approximately normally distributed. Answer: False Difficulty: Medium 5.
* Its benefits are price it’s less expensive to build and run. But if one of the lines goes down then your whole network goes down. Exercise 7.2.4 Circuit classification | Total Bandwidth (Mbps) | Features and summary | T1 | 1.544 | Uses 24 channels to communicate. | E1 | 2.048 | Uses 32 channels to communicate. | T3 | 44.7 | Uses 672 channels to communicate.
| | | |X |Are the inputs to the program identified? | | | |X |Are the outputs to the program identified? | | | |X |Does the test input result in the expected output? | | List of Defects Identify defects by checklist number and observed error, but do not include any corrective action. Display Menu Test Test #1: Valid Menu Selection = Canadian Dollars Inputs: Menu Selection = 1 Expected Output: “Do you want to continue with the conversion, Y = Yes, N = No” Currency Type = 1 Test #2: Invalid Menu Selection = Error Inputs: Menu Selection = 7 Expected Output: “Error 4: Invalid menu selection” Start
Practice Test 4 –Bus 2023 Directions: For each question find the answer that is the best solution provided. There is only one correct answer. 1. For a sample size of 30, changing from using the standard normal distribution to using the t distribution in a hypothesis test, a. will result in the rejection region being smaller b. will result in the rejection region being larger c. would have no effect on the rejection region d. Not enough information is given to answer this question. ANSWER: a -recall that all the t-values are larger than the z-values so it makes it more difficult to reject the null.
To multiply and divide radicals, perform the indicated operation on the radicands and then simplify: (You do not need radicand to be the same in order to multiply or divide) Example 1: (85) (42) = (8)(4) 5 2 = 3210 Example 2: (105) ÷ (25)= (10 ÷ 2) (5 ÷ 5) = 5 (1) = 5 Example 3: (320 ) (32) sometimes it is easier to simplify first (64 ∙ 5) (16 ∙ 2 (85) (42) Answer = 3210 Example 4: (6500) ÷ 220 ← Rewrite as a fraction and divide =
Marcus Davis May 14, 2014 MAT 220 Week 2 40.) What does it means if two quantities vary directly? It means that if one variable decreases or increases than the other variable will vary to the same exact degree. 41.) In your own words explain how to solve a variation problem.
It is reasonable to set the value to 0 because if one makes the value 0 we can observe a more precise answer, than if one changes the intercept. When I changed the intercept to 1 the equation decreased and had a +1 to it. It slowly kept decreasing as I increased the number. 6) Compare the equation of the line of best fit for your random sample with the lines