Absolute value (ABSO) is: Represents the distance a number is from zero no matter it’s a positive or negative number. For example, the number 5 would have an absolute valve of 5 and the number -5 would also have an absolute value of 5. 11. Define Confidence interval and level In statistics, a confidence interval (CI) is a type of interval estimate of a population parameter and is used to indicate the reliability of an estimate. It is an observed interval in principle different from sample to sample, that frequently includes the parameter of interest if the experiment is
3. Be able to conduct a goodness of fit test when the population is hypothesized to have a multinomial probability distribution. 4. For a test of independence, be able to set up a contingency table, determine the observed and expected frequencies, and determine if the two variables are independent. 5.
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.
The physically impossibility of checking all the items in the population C. The adequacy of sample results D. All the above are reasons for sampling 5. Which of the following types of samples can you use if you want to make valid statistical inferences from a sample to a population? A. A judgment sample B. A quota sample C. A convenience sample D. A probability sample 6.
Which of the following tests could he use for this? a. An independent samples t-test. b. ANOVA c. A paired-samples t-test. *d. A Chi-square test.
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
In what class interval must the median lie? Explain your answer. (5 pts) 5. Assume that the largest observation in this dataset is 5.8. Suppose this observation wereincorrectly recorded as 8.5 instead of 5.8.
For what values of t will the null hypothesis not be rejected? a) To the left of -1.645 or to the right of 1.645 b) To the left of -1.345 or to the right of 1.345 c) Between -1.761 and 1.761 d) To the left of -1.282 or to the right of 1.282 QNT 561 Final Questions and Answers QNT 561 Final Exam 2. Which of the following is a characteristic of the F distribution? a) Normally distributed b) Negatively skewed c) Equal to the t-distribution d) Positively skewed 3. For a chi-square test involving a contingency table, suppose the null hypothesis is rejected.
Asking the question, ‘what do I observe?”gives the dependent variable the stability of the predicted outcomes for the research that is researched. Independent Variables By choosing five [demographic, give and take, responsiveness, trust, and fairness] of the twelve independent variables, a parsimonious effect is given because it explains the most with the least amount of independent variables. These variables influence the dependent variable by asking the question, “what do I change?’ and can predict the dependent variable with cause and effect of the research outcomes. These five selected variables can form a relationship with another chosen variable to establish a standardized measure of strength or a Pearson’s correlation coefficient (see table 2). [Insert Table 2 here] Central Tendency The three measurements used to calculate the frequency distribution is the mean, median, and mode.
Statistics 121 Problem set #1 1. Suppose {A, B, C, D, E, F} is a partition of the sample space Ω. Suppose it is known that P(A∪B∪C)= 0.6. The probability of event B is the same as the probability of event D. It is also known that P(A∪B)= P(E∪F)= 0.5 P(B∪C). Find the probabilities of the following events: a) B Solution: PA∪B∪C=0.6 ; PD∪E∪F=0.4 PB= PD = PA∪B= PE∪F = PA+ PB= PE+ PF = PA+ PB= 0.4- PD = PA+ PB= 0.4- PB = PA+ 2PB= 0.4 PA∪B= 0.5PB∪C 2PA+ 2PB= PB+ PC 0.4 + PA= PB+ PC 0.4 + PA+PA= PA+PB+ PC 0.4 + 2PA= 0.6 2PA= 0.2 PA= 0.1 PA+ 2PB= 0.4 0.1 + 2PB= 0.4 2PB=0.3 PB=0.15 b) c) A∪B∪D = PA+ PB+ PD = PA+ 2PB = 0.1 + 2(0.15) P(A∪B∪D)= 0.4 * I can’t find numbers 2, 3 and 4 5.