Daniel Jones NT1210 Lab 1.1 Review 1. Convert the decimal value 127 into binary. Explain the process of conversion that you used. 127 | 127 | 63 | 31 | 15 | 7 | 3 | 1 | 128 | - 64 | - 32 | - 16 | - 8 | - 4 | - 2 | - 1 | | = 63 | = 31 | = 15 | = 7 | = 3 | = 1 | = 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | The answer is: 01111111 If the decimal number is less than the greatest power of 2 than you must put a 0 for that number than carry that same decimal number over to the right one decimal place. For example.
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.
My hypothesis for numbering of fruits was that Vita Lite was going to have more fruits by a good amount but the normal lighting did end up producing more druits maybe due to brighter conditions it was able to simulate. The vita lite seemed very dull. {draw:frame} Figure 5 explains the germination of the beginning of the expierment how well the seeds germinated. My Hypothesis was undecided because I knew for germination neither of the different lights were going to make a diffrence with the germination. Due to it only needs to be watered and get some nutrition.
A trend line is a line that shows the general trend of the data and, therefore, does not have to touch each point. Answer: (3 points) Score 6. Compare your data with your hypothesis. Explain how the data supports or does not support your hypothesis. If your data does not support your hypothesis, use the data you collected to answer the original question.
(b) One correct answer to this part is as follows: We can say that we are 95 percent confident the average balance for all overdue bills is in the range from $480 to $520. Since the standard error SE for the sample mean should equal 100/square root of 100 = 10, the margin of error would be twice this value, or $20. This means $480 to $520 is a 95 percent confidence interval for the average of all overdue balances. Another correct answer to this part is as follows: Instead of the average balance, we can find the percentage of all balances that fall in this interval. When we think of all balances, $480 is 0.2 standard deviations below the average (480 -500)/100 = - 0.20), and $520 is 0.2 standard deviations above the average (520 -500)/100 = 0.20).
5. A test statistic is a value determined from sample information used to reject or not reject the null hypothesis. 6. The region or area of rejection defines the location of all those values that are so large or so small that the probability of their occurrence under a true null hypothesis is rather remote. 7.
Type Your Name in the Report title. C. Page numbers should be RIGHT JUSTIFIED at the page bottom. D. Ensure you have at least 1” spacing all around. 1. Create a report that Sum Sales, Costs, and Profit by Region.
Sample: I-2B Score: 5 Part (a)(i): 2 points were earned for the correct setup and the correct answer. Part (a)(ii): 1 point was earned. The student makes computational errors that limit the score. Part (b): No points were earned. The setup is nearly correct, but the computation of 10 m3 of infiltration into the surrounding soil should equal 100 m3.
1. A random sample of size 15 is selected from a normal population. The population standard deviation is unknown. Assume the null hypothesis indicates a two-tailed test and the researcher decided to use the 0.10 significance level. For what values of t will the null hypothesis not be rejected?
Assignment 3 (2014) QA026 1. A die is biased so that, when it is rolled, the probability of obtaining a score of 6 is ¼. The probabilities of obtaining each of the five scores 1, 2, 3, 4, 5 are equal. Calculate the probability of obtaining a score of five with this biased die. a.