University:
Carleton College
Course:
MATH 111 | lntroduction to Calculus
Academic year:

2024
Views:

139

Pages:

3
Author:

Alecai8y

Question 1 In a normal distribution, a data value located 1 standard deviations below the mean has Standard Score: z = In a normal distribution, a data value located 2.2 standard deviations above the mean has Standard Score: z = In a normal distribution, the mean has Standard Score:z Answer 1-x z = σ The z-score indicates how many standard deviations above or below the mean a raw score or data item falls. The z-numerical score's value indicates the departure of the data value from the mean in units of standard deviations, and the z-sign score's indicates the direction with regard to the mean. The z-score is positive if the data value is greater than the mean. The z-score is negative if the data value is below the mean. The data value in this case is said to be one standard deviation from the mean. The phrase "I standard deviation" denotes that the z-value score's must be 1, whereas the phrase "below" denotes that its sign must be negative. b. The data value is said to be 2.2 standard deviations above the mean. The required z-score is 2.2. χ- μ C. z = σ μ-μ = 0. σ σ The required z-score for the mean is 0. Question 2 What are the data modeling concepts used in MongoDB? What are the main CRUD operations of MongoDB? Answer Dato modeling concept: Documets are stored in BSON(binary JSON) fermat, more efficient than JSON. Individual documents are stored in a collection. Collection has no schema. Mongo Db has several Chub operations, where CRUD stands for(create, read update, delete). Insert operation, whose format is: df . insert (cdocument(s) >) Remove operation and the format is: db. .remove() For read queries, the main command is called find, and the format is: dl. .find(condition) Question 3 Give a real-life example of data which could have the following probability distributions Solve a) bimodal distibution b) uniform distribution c) exponential distribution only early Answer b). Uniform distribution-: The uniform distribution occurs when all conceivable outcomes are equally likely to occur. The real-life illustration of a uniform distribution is the act of drawing a card from a deck. c). Exponential distribution: An exponential distribution is a data set in which the events are distributed over time (or by time). For the real-life example, The time that until an accident occurs, is the real-life example of an exponential distribution. a) Bimodal distribution-: The distribution of data set with two nodes is known as the Bimodal distribution. For the real-life example, The number of passengers in Delhi Metro in a day (except weekends), in the morning and evening, the number of passengers is creating two nodes, is the real-life example of the bimodal distribution. Question 4 Use the change-of-base theorem to find an approximation to four decimal places for each logarithm log25 Answer Given, log25 Change to base 10, we get log 5 logax log25 = :'logx = log 2 logab 0.69897 0.0301029 = 2.3219 Therefore, log25 = 2.3219 Question 5 When is the chi squared statistic employed for hypothesis testing? When testing a hypothesis on a single population, be sure to include supporting assumptions and the test statistic. What standard must be met in order to reject the null hypothesis in both directional and non-directional tests? In each situation, how do you determine the p value? Answer Chi-square statistic for testing single variance: To determine the population variance of a single sample, use the Chi-square statistic. The necessary assumptions for Chi-square test: Simple random sampling should be used to obtain the sample. The sample's source population ought to exhibit normal distribution. The data should be continuous. The results of the chi-square test are as follows: x2 = (n-1)s² σ2 n = Sample size s2 = Sample variance 0² = Population variance Decision rule based on P-value approach for both directional and non-directional tests: The level of significance is a. If P-value ≤ a, then reject the null hypothesis Ho. If P-value > a, then fail to reject the null hypothesis Ho. P-value: The P-value will be obtained from the chi-square distribution table based on the value of test statistic and the degrees of freedom (n-1) and the type of hypothesis test (Two tailed, right tailed or left tailed). Chi-square statistic for testing distribution: To determine whether the sample data are consistent with the hypothesized distribution or not, the Chi-square goodness of fit test is used. The chi-square goodness-of-fit test is used to test whether a sample of data comes from a population with a specific distribution. The chi-square goodness-of-fit can also be applied to discrete distributions The necessary assumptions for Chi-square test for goodness of fit are given below: The sample should be collected using simple random sampling. The variable of interest must be categorical. The expected value of each cell should not be less than 5. Evidently, the test is to determine whether a sample of data comes from a population with a specific distribution. Chi-square goodness of fit is a right tailed test. Therefore, it is a directional test. The chi-square test statistic is obtained as given below: x2 = Σ−1 (Ο - Εί)2 Ei 0₁ = Observed frenquency E₁ = Expected frequency Decision rule based on P-value approach: The level of significance is a. If P-value ≤ a, then reject the null hypothesis Ho. If P-value > a, then fail to reject the null hypothesis Ho. The P-value will be obtained from the chi-square distribution table based on the value of test statistic and the degrees of freedom (n-1) for the right tailed test.

Modeling data distributions #3 - Questions and Answers

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