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Hypothesis-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 pro…

Hypothesis-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…

Hypothesis 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

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

Hypothesis-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 variable…

Hypothesis-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.

Hypothesis 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…

Chi2 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.

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 a…

Hypothesis-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. De…

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

Data Science - Hypothesis Testing Work

Testing the hypothesis of a pharma company for the time of the effect and the quality assurance based on the null and alternate hypothesis

Chi2 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

Used libraries and functions as follows: