alternate-hypothesis
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tantawy997 / Analyze_ab_test_results_notebook
Analyze ab test results udacity project
Language:Jupyter Notebook 2aldimeolaalfarisy / Hypothesis-Testing-Concept-ANOVA-
ANOVA test using python to find out if survey or experiment results are significant and the impact of one or more factors by comparing the means of different samples
Language:Jupyter Notebook 1vaitybharati / Assignment-03-Q4-Hypothesis-Testing-
Chi2 contengency independence test. Q4. TeleCall uses 4 centers around the globe to process customer order forms. They audit a certain % of the customer order forms. Any error in order form renders it defective and has to be reworked before processing. The manager wants to check whether the defective % varies by centre. Please analyze the data at 5% significance level and help the manager draw appropriate inferences.
Language:Jupyter Notebook 1Chi2 contengency independence test. Fantaloons Sales managers commented that % of males versus females walking in to the store differ based on day of the week. Analyze the data and determine whether there is evidence at 5 % significance level to support this hypothesis.
Language:Jupyter Notebook 1Hypothesis Testing 1S2T - Call Center Process. Sample Parameters: n=50, df=50-1=49, Mean1=4, SD1=3 1-sample 2-tail ttest Assume Null Hypothesis Ho as Mean1 = 4 Thus, Alternate Hypothesis Ha as Mean1 ≠ 4
Language:Jupyter Notebook 1Hypothesis-Testing-1-Sample-1-Tail-Test-Salmonella-Outbreak. 1-sample 1-tail ttest. Assume Null Hypothesis Ho as Mean Salmonella <= 0.3. Thus Alternate Hypothesis Ha as Mean Salmonella > 0.3. As No direct code for 1-sample 1-tail ttest available with unknown SD and arrays of means. Hence we find probability using 1-sample 2-tail ttest and divide it by 2 to get 1-tail ttest.
Language:Jupyter Notebook 1Hypothesis-Testing-2-Sample-2-Tail-Test-Drugs-and-Placebos. Note: This python code states both 2-sample 1-tail and 2-sample 2-tail codes. Treatment group mean is Mu1 Contrl group mean is Mu2 2-sample 2-tail ttest Assume Null Hypothesis Ho as Mu1 = Mu2 Thus Alternate Hypothesis Ha as Mu1 ≠ Mu2.
Language:Jupyter Notebook 1Hypothesis-Testing-2-Proportion-T-test-Students-Jobs-in-2-States. Assume Null Hypothesis as Ho is p1-p2 = 0 i.e. p1 ≠ p2. Thus Alternate Hypthesis as Ha is p1 = p2. Explanation of bernoulli Binomial RV: np.random.binomial(n=1,p,size) Suppose you perform an experiment with two possible outcomes: either success or failure. Success happens with probability p, while failure happens with probability 1-p. A random variable that takes value 1 in case of success and 0 in case of failure is called a Bernoulli random variable. Here, n = 1, Because you need to check whether it is success or failure one time (Placement or not-placement) (1 trial) p = probability of success size = number of times you will check this (Ex: for 247 students each one time = 247) Explanation of Binomial RV: np.random.binomial(n=1,p,size) (Incase of not a Bernoulli RV, n = number of trials) For egs: check how many times you will get six if you roll a dice 10 times n=10, P=1/6 and size = repetition of experiment 'dice rolled 10 times', say repeated 18 times, then size=18. As (p_value=0.7255) > (α = 0.05); Accept Null Hypothesis i.e. p1 ≠ p2 There is significant differnce in population proportions of state1 and state2 who report that they have been placed immediately after education.
Language:Jupyter Notebook 1Hypothesis Testing Anova Test - Iris Flower dataset. Anova ftest statistics: Analysis of varaince between more than 2 samples or columns. Assume Null Hypothesis Ho as No Varaince: All samples population means are same. Thus Alternate Hypothesis Ha as It has Variance: Atleast one population mean is different. As (p_value = 0) < (α = 0.05); Reject Null Hypothesis i.e. Atleast one population mean is different Thus there is variance in more than 2 samples.
Language:Jupyter Notebook 1Hypothesis-Testing-Chi2-Test-Athletes-and-Smokers. Assume Null Hypothesis as Ho: Independence of categorical variables (Athlete and Smoking not related). Thus Alternate Hypothesis as Ha: Dependence of categorical variables (Athlete and Smoking is somewhat/significantly related). As (p_value = 0.00038) < (α = 0.05); Reject Null Hypothesis i.e. Dependence among categorical variables Thus Athlete and Smoking is somewhat/significantly related.
Language:Jupyter Notebook 1Hypothesis-Testing-Chi2-Test-Human-Gender-and-Choice-of-Pets. Assume Null Hypothesis as Ho: Human Gender and choice of pets is independent and not related. Thus Alternate Hypothesis as Ha : Human Gender and choice of pets is dependent and related. As (p_valu=0.1031) > (α = 0.05); Accept Null Hypothesis i.e Independence among categorical variables. Thus, there is no relation between Human Gender and Choice of Pets.
Language:Jupyter Notebook 1saikrishnabudi / Hypothesis-Testing
Data Science - Hypothesis Testing Work
Language:Jupyter Notebook 0shwetapardhi / Assignment-03-Q4-Hypothesis-Testing
Chi2 contengency independence test Q4. TeleCall uses 4 centers around the globe to process customer order forms. They audit a certain % of the customer order forms. Any error in order form renders it defective and has to be reworked before processing. The manager wants to check whether the defective % varies by centre. Please analyze the data at 5
Language:Jupyter Notebook 0Chi2 contengency independence test Q5. Fantaloons Sales managers commented that % of males versus females walking in to the store differ based on day of the week. Analyze the data and determine whether there is evidence at 5 % significance level to support this hypothesis. Assume Null Hypothesis as Ho: Independence of categorical variables (% of
Language:Jupyter Notebook 0Lakshya-Ag / Inferential-Statistics-and-Hypothesis-Testing
Testing the hypothesis of a pharma company for the time of the effect and the quality assurance based on the null and alternate hypothesis
PatilSukanya / Assignment-3-Hypothesis-Testing
Used libraries and functions as follows:
Language:Jupyter Notebook